Last week, we saw how to reduce the quartic equation to the cubic equation. This was done by carefully analyzing the symmetries of the solutions to the equation. Today, our question is: how do we solve cubic equations -- for example, what are the solutions to \[ x^3+6x^2+9x+3=0? \] In mathematics, when you cannot solve a problem, it is helpful to compare it to a problem you already know how to solve. So, we are going to start by revisiting the quadratic formula.
By this simple change of variables \(y = x+h,\) we were able to get rid of the linear term in our equation! This is the power of completing the square.
From here, getting the quadratic formula is easy: taking a square root, and using that \(x = y-h,\) we find \[x = -\frac{b}{2} \pm \sqrt{\frac{b^2}{4} - c}.\]
Let's now review the approach to the quadratic formula we used last week.
Denote the two solutions of the quadratic equation \(x^2+bx+c=0\) by \(\alpha\) and \(\beta.\) Then \[x^2 + bx + c = (x-\alpha)(x-\beta),\] so Vieta's formulas tell us \[\alpha + \beta = -b\] and \[\alpha\beta = c.\] As we saw, any symmetric polynomial of \(\alpha\) and \(\beta\) can be rewritten in terms of \(\alpha+\beta\) and \(\alpha\beta,\) and hence in terms of \(b\) and \(c\) -- for example, \[\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = b^2 - 2c,\] so even if we haven't solved our quadratic yet, we can still evaluate \(\alpha^2 + \beta^2\) just using the coefficients of \(x^2 + bx + c = 0.\)
To solve for \(\alpha, \beta,\) we needed to find a way to evaluate some quantity which wasn't symmetric in \(\alpha\) and \(\beta.\) The symmetry-breaking quantity we chose was \(\alpha - \beta.\) This is not symmetric in \(\alpha\) and \(\beta\), as swapping the two roots negates this expression. However, it's not so far from being symmetric: \[(\alpha - \beta)^2\] is symmetric in \(\alpha, \beta,\) and so we can write \((\alpha-\beta)^2\) in terms of \(b\) and \(c\); explicitly, we found \[(\alpha-\beta)^2 = b^2 - 4c.\] Taking a square root then gives us the asymmetric quantity \[\alpha - \beta = \sqrt{b^2 - 4c}.\]
But, in this derivation, how does one come up with the idea to use \(\alpha - \beta\)? One way to motivate this is that we know the quadratic formula is going to involve a square root, and \(\alpha - \beta\) has the same symmetry as the square root function. What do we mean by this?
Square roots always come in positive/negative pairs: the equation \(y^2 = 4\) has solutions \(y = 2\) and \(y = -2,\) for example.
In our final square root formula, we wrote \[\alpha - \beta = \sqrt{b^2 - 4c}.\] The positive/negative ambiguity in the square root on the right hand side is reflected in the positive/negative ambiguity in the expression \(\alpha-\beta\) on the left hand side: depending on which root we call \(\alpha\) and which we call \(\beta,\) the expression \(\alpha - \beta\) could take on two possible values, which are negatives of each other, just like how the square root can be either positive or negative.
We now want to derive a cubic formula, to solve equations \[x^3 + ax^2 + bx + c = 0.\]
Before doing anything with symmetries of numbers, we can ask ourselves: does completing the square work here? That is, can we simplify a cubic equation by doing a substitution of the form \(y = x + h\), for some well-chosen \(h\)?
It turns out that one can always reduce a cubic equation to one of the form \[y^3 + py + q = 0,\] just by choosing \(h\) right.
If we substitute \(y = x+h,\) then the equation \[x^3 + ax^2 + bx + c = 0\] becomes \[(y-h)^3 + a(y-h)^2 + b(y-h) + c = 0,\] or equivalently \[y^3 + (a - 3h)y^2 + (b - 2ah + 3h^2)y + (c - bh + ah^2 - h^3) = 0.\]
In particular, if we choose \(h = a/3,\) then our substitution \(y = x + a/3\) turns \[x^3 + ax^2 + bx + c = 0\] into an equation of the form \[y^3 + py + q = 0,\] as desired.
Cubic equations of the form \[y^3 + py + q = 0\] are typically called depressed cubics. Unfortunately, the depressed cubic still has this pesky \(py\) term, but it's a bit easier than the general cubic, so we will work with it in the rest of the article to make the algebra easier. As we just saw, we're not losing any generality: any cubic can be turned into a depressed cubic, so solving depressed cubics gives you a way to solve all cubics.
So, take the depressed cubic \[y^3 + py + q = 0,\] and label its roots \(y = \alpha\), \(y = \beta,\) and \(y = \gamma.\) We will now find an equation for these roots.
Now that completing the square has given us as much help as it can, we turn to the symmetries of numbers.
The cubic formula will have to involve the cube root function. However, it will also involve square roots. To see why, notice that we can turn any quadratic equation into a cubic equation: the equation \(y^2 - 2 = 0\) can be multiplied by \(y,\) transforming it into the depressed cubic \[y^3 - 2y = 0.\]
The solutions to this cubic are \(y=0\), \(y=\sqrt{2},\) and \(y=-\sqrt{2}.\) So, if we are going to have a general cubic formula, it will need to involve square roots.
In the quadratic case, we figured out what quantity to take the square root of by trying to find some quantity which was multiplied by \(-1\) when we permuted the roots. Our game plan for the cubic will be to first figure out what quantity we should take the square root of, then figure out what we will take the cube root of.
To find the square root part in the cubic formula, we need to find some function of \(\alpha, \beta,\) and \(\gamma\) that isn't symmetric, but only in a mild way: relabelling the roots should at worst replace this quantity with its negative.
To start, let's copy what we did for the quadratic and consider the quantity \(\alpha - \beta\) and see what happens. When we swap the values of \(\alpha\) and \(\beta,\) this becomes \(\beta - \alpha,\) the negative of itself. So, we're off to a good start!
But now that we have three roots, we have six different ways to permute them.
For example, the cubic \(y^3 - 3y + 1 = 0\) has solutions \(y \approx -1.88, y \approx 0.35, y \approx 1.53.\) Then, there are six ways of deciding how to call these roots \(\alpha,\beta,\) and \(\gamma.\)
| \( \alpha = -1.88,\\ \beta = 0.35,\\ \gamma = 1.53 \) | \( \alpha = -1.88,\\ \beta = 1.53,\\ \gamma = 0.35 \) | \( \alpha = 0.35,\\ \beta = -1.88,\\ \gamma = 1.53 \) |
| \( \alpha = 0.35,\\ \beta = 1.53,\\ \gamma = -1.88 \) | \( \alpha = 1.53,\\ \beta = -1.88,\\ \gamma = 0.35 \) | \( \alpha = 1.53,\\ \beta = 0.35,\\ \gamma = -1.88 \) |
And while \(\alpha - \beta\) transforms well when we swap \(\alpha\) and \(\beta,\) if we had instead swapped \(\beta\) and \(\gamma\) then something bad would happen: \(\alpha - \beta\) would become \(\alpha - \gamma,\) which is completely unrelated to \(\alpha - \beta.\)
Fortunately, there's a fix. Consider the product \[D = (\alpha - \beta)(\beta - \gamma)(\gamma - \alpha).\]
If we swap \(\alpha\) and \(\beta,\) then this turns into \[(\beta - \alpha)(\alpha - \gamma)(\gamma - \beta).\] Notice that this new product is equal to \((-1)^3 \cdot D = -D,\) because if we negate each factor of this product, we get \(D.\)
Similarly, if we swap \(\alpha\) and \(\gamma\) or if we swap \(\beta\) and \(\gamma,\) then \(D\) becomes \(-D.\) We're off to a good start!
Those `root swaps' correspond to three different labellings of our roots; if we count the original \(\alpha,\beta,\gamma\) way, then that's four permutations accounted for. How about the last two?
We could also cycle the values; for example, going back to that cubic \(y^3 - 3y + 1 = 0\) from above, if we started with \[\alpha = -1.88,\ \beta = 0.35,\ \gamma = 1.53,\] then we could move each value forward one to relabel them as \[\alpha = 1.53,\ \beta = -1.88,\ \gamma = 0.35.\] Notice how the new value of \(\alpha\) is the old \(\gamma\); the new value of \(\beta\) is the old \(\alpha\); and the new \(\gamma\) is the old \(\beta.\) Thus, under this type of permutation, \(D = (\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)\) becomes \[(\gamma - \alpha)(\alpha - \beta)(\beta - \gamma),\] which is equal to \(D\) itself!
Similarly, we could cycle the values the other way: instead of labelling the roots \[\alpha = -1.88,\ \beta = 0.35,\ \gamma = 1.53,\] we could label them as \[\alpha = 0.35,\ \beta = 1.53,\ \gamma = -1.88.\] Under this relabelling, \(D\) also is unchanged!
We now have this quantity \(D\) which has the property that, when we permute the values \(\alpha, \beta,\) and \(\gamma,\) the quantity \(D\) is either unchanged or multiplied by \(-1.\)
In particular, \(D^2\) is a symmetric function of \(\alpha,\beta,\) and \(\gamma\) -- if we relabel the values of \(\alpha,\beta,\gamma,\) then \(D^2\) always stays the same, because the relabelling can at worst change us from \(D^2\) to \((-D)^2,\) but \(D^2 = (-D)^2.\)
Thus, as we discussed last week, the symmetric function \(D^2\) can be written in terms of the coefficients \(p,q\) of our cubic equation \[y^3 + py + q = 0.\]
If you go through the algebra to do this explicitly, you can compute \[D^2 = -4p^3 - 27q^2.\]
Recall from Vieta's formulas (discussed last week) that \[\alpha + \beta + \gamma = 0,\] \[\alpha\beta + \beta\gamma + \gamma\alpha = p,\] \[\alpha\beta\gamma = -q.\]
We know \(D^2=(\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2\) is some polynomial \(P(p,q)\) of \(p\) and \(q\). Our task is to compute \(P(p,q)\).
Let's substitute \(\gamma=0\), which implies \(\beta=-\alpha\) and \[q=0,\ \ p=-\alpha^2.\] This gives \[ P(p,0)=D^2=(2\alpha)^2(-\alpha)^2(-\alpha)^2=4\alpha^6=-4p^3. \] This means \(P(p,q)\) must contain a term \(-4p^3\).
Now, we may compute \[P(p,q)+4p^3=(\alpha-\beta)^2(\beta-\gamma)^2(\gamma-\alpha)^2+4(\alpha\beta+\beta\gamma+\gamma\alpha)^2=27\alpha^2\beta^2\gamma^2.\] The last equality can be checked for example by substituting \(\gamma=-\alpha-\beta\) to both sides to get rid of one variable.
We conclude that \(P(p,q)=-4p^3-27q^2\).
The upshot to all this is that we should expect the cubic formula will involve the quantity \[\sqrt{-4p^3 - 27q^2}.\]
There are two possible choices of that square root; they correspond to the two possible values of \(D\) (this \(D\) has two values because, depending on how you label the roots \(\alpha, \beta,\) and \(\gamma,\) this \(D\) could take on two opposite values).
In particular, if we choose one of the values of that square root, then we eliminate some symmetry. This is just like what happened when we chose a square root when solving the quadratic case: if we choose a square root, and set \(D\) equal to that square root, then we can no longer swap two roots, because swapping two roots would negate our value of \(D,\) but now we've fixed \(D.\) Observe how by choosing \(D,\) we've reduced the symmetry.
However, we haven't gotten rid of all the symmetries: while the root swapping symmetries go away, above we saw that those two `cycle the roots' symmetries don't change \(D\) at all. Thus, we now have three valid permutations of \(\alpha, \beta, \gamma\) remaining.
The cubic formula also has to involve a cube root. So, let's ask ourselves: which quantity should we be taking the cube root of?
Well, let's take a step back. It is well known that square roots come in pairs. But how many solutions to \(z^3 = 1\) are there, say?
We can rewrite this equation as \[z^3 - 1 = 0,\] and then factor it as \[(z-1)(z^2 + z + 1) = 0.\]
The quadratic equation \(z^2 + z + 1 = 0\) has solutions \[z = \frac{-1 \pm \sqrt{-3}}{2}.\]
People often use the symbol \(\omega\) (omega) to denote the number \[\omega = \frac{-1 + \sqrt{-3}}{2}.\] Notice that \(\omega\) is a complex number, because \(\sqrt{-3}\) is imaginary. For those who know Euler's formula, we remark that \(\omega = e^{2\pi i/3}.\)
The other root of \(z^2 + z + 1 = 0\) is simply \(\omega^2\) (check for yourself!). Thus \(z^3 = 1\) has three solutions: \(1, \omega,\) and \(\omega^2.\) This means that every cube root problem has three solutions. For instance, \(2^3 = 8,\) but \(8\) also has two more cube roots: \[(2\omega)^3 = 2^3 \cdot \omega^3 = 8 \cdot 1 = 8\] and \[(2\omega^2)^3 = 2^3 \cdot (\omega^2)^3 = 8 \cdot 1 = 8.\]
So, while square roots come in pairs, cube roots come in triples \(r, r\omega, r\omega^2\).
When solving the quadratic equation, we tried to find some quantity which had the same symmetry as the square root function. Now, we want to find some expression in terms of \(\alpha, \beta, \gamma\) (the roots of our cubic) which has the symmetry, that when you permute the roots, the expression is multiplied by \(\omega\) or \(\omega^2.\)
When we wanted an expression which was multiplied by \(-1\) when we swapped the roots, we used \(\alpha + (-1)\cdot \beta.\) This inspires us to use the following trick: consider \[A = \alpha + \omega \cdot \beta + \omega^2 \cdot \gamma.\]
Recall the two cycling symmetries from above, where we cycle all the values around either left or right; under these cycling symmetries, the quantity \(A\) above transforms to either \[\gamma + \omega \cdot \alpha + \omega^2 \cdot \beta\] or \[\beta + \omega \cdot \gamma + \omega^2 \cdot \alpha,\] depending on which way you cycle the values.
But recall that \(\omega^3 = 1,\) so that \[\omega \cdot A = \gamma + \omega \cdot \alpha + \omega^2 \cdot \beta\] and \[\omega^2 \cdot A = \beta + \omega \cdot \gamma + \omega^2 \cdot \alpha.\]
This is great: it means that, under these cyclicing permutations, the quantity \(A\) gets multiplied by \(\omega\) or \(\omega^2\) -- this is exactly the way we'd hope a cube root gets changed! In particular, \(A^3\) is completely unchanged by the two cycling permutations.
The swapping permutations (where we swap the value of \(\alpha\) and \(\beta\) and leave \(\gamma\) alone, for example) don't respect \(A\) at all, though: they send \(A\) -- and hence also \(A^3\) -- to completely different values.
If \(A^3\) were completely symmetric, then we could express it in terms of \(p\) and \(q.\) However, one of key insights of Galois theory is that we can exploit the partial symmetry of \(A^3\) to rewrite it in terms of \(p, q,\) and \(D\).
Recall, in the last section, that choosing a value of \(D\) breaks the swapping symmetries: once we choose \(D,\) we've eliminated the swapping symmetry. As the swapping symmetry is the only reason that \(A^3\) isn't perfectly symmetric, we can hope that we can express \(A^3\) in terms of \(p\) and \(q\) (symmetric quantities), together with \(D\) (another quantity which breaks symmetry in the exact same \(A^3\) does).
With a healthy amount of algebra, we can realize this goal, and write \[A^3 = -\frac{27}{2}q - \frac{3i\sqrt{3}}{2}D.\]
We can expand out \[A^3 = (\alpha + \omega\beta + \omega^2\gamma)^3\] to get the rather lengthy expression \[A^3 = (\alpha^3 + \beta^3 + \gamma^3+ 6\alpha\beta\gamma) + 3\omega(\alpha^2\beta + \beta^2\gamma + \gamma^2\alpha) + 3\omega^2(\alpha^2\gamma + \gamma^2\beta + \beta^2\alpha).\] This expression is still rather complicated, so let's deal with it piece by piece. At the start of \(A^3,\) we find the completely symmetric quantity \[\alpha^3 + \beta^3 + \gamma^3+6\alpha\beta\gamma.\]
Let's first compute \(\alpha^3 + \beta^3 + \gamma^3\). Since our cubic equation has no quadratic term, Vieta's formulas tell us \[\alpha+\beta+\gamma=0.\] Thus we have \(\gamma=-\alpha-\beta,\) and hence we can rewrite \(\alpha^3+\beta^3 + \gamma^3\) as \[ \alpha^3 + \beta^3 + \gamma^3=\alpha^3+\beta^3+(-\alpha-\beta)^3=-3\alpha^2\beta-3\alpha\beta^2. \] \[ -3\alpha^2\beta-3\alpha\beta^2=3\alpha\beta(-\alpha-\beta). \] We can again replace \(-\alpha-\beta\) by \(\gamma\), which means the equation equals \[ 3\alpha\beta\gamma. \] Finally by Vieta's formula we know \(q=-\alpha\beta\gamma\), which tells us \[ \alpha^3 + \beta^3 + \gamma^3 = -3q. \] We can add \(6\alpha\beta\gamma=-6q\) to this to get \[\alpha^3 + \beta^3 + \gamma^3 +6\alpha\beta\gamma= -9q.\]
Therefore \[A^3 = -9q + 3\omega(\alpha^2\beta + \beta^2\gamma + \gamma^2\alpha) + 3\omega^2(\alpha^2\gamma + \gamma^2\beta + \beta^2\alpha).\]
Set \[u = \alpha^2\beta + \beta^2\gamma + \gamma^2\alpha,\] \[v = \alpha^2\gamma + \gamma^2\beta + \beta^2\alpha\] so that \[A^3=-9q+3\omega u+3\omega^2v.\] We now rewrite both of \(u, v\) in terms of \(p, q,\) and \(D.\)
As \(u\) and \(v\) are not symmetric with respect to swapping permutations, we will really need to use \(D\) to express them. So, let's think about \[D = (\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)\] for a moment. If we expand it out, we get \[D = \alpha^2(\gamma - \beta) + \beta^2(\alpha - \gamma) + \gamma^2(\beta - \alpha),\] so that, remarkably, \[D = v - u.\]
Observe also that, while \(u\) and \(v\) are individually asymmetric, the sum \(u+v\) is symmetric: \[u + v = \alpha^2(\beta + \gamma) + \beta^2(\alpha + \gamma) + \gamma^2(\alpha + \beta)\]
As \(\alpha + \beta + \gamma = 0,\) we can further rewrite \(u+v\) as \[u+v = \alpha^2(-\alpha) + \beta^2(-\beta) + \gamma^2(-\gamma)=-\alpha^3 - \beta^3 - \gamma^3.\] But a moment ago, we computed \(\alpha^3 + \beta^3 + \gamma^3 = -3q,\) so that \[u+v = 3q.\]
Thus we have two variables \(u\) and \(v,\) and two linear equations: \(v - u = D\) and \(v + u = 3q.\) We can solve this system to find \[v = \frac{D + 3q}{2},\] \[u = \frac{3q - D}{2}.\]
Therefore \[A^3 = -9q + \frac{3\omega}{2}(3q - D) + \frac{3\omega^2}{2}(3q + D),\] or even more simply \[A^3 = -\frac{27}{2}q - \frac{3i\sqrt{3}}{2}D,\] as desired.
In particular, if we choose a cube root root of \[-\frac{27}{2}q - \frac{3i\sqrt{3}}{2}D,\] then we'll have a value for \(A.\) Thus, we now know the value of \[A = \alpha + \omega\beta + \omega^2\gamma.\] We also know that \(\alpha + \beta + \gamma = 0.\) We have three variables, and two linear equations. We can almost solve for \(\alpha,\beta,\gamma\) easily, but we need one more linear equation between them to do it (if you have \(n\) linear equations in \(n\) variables, it's also easy to solve for the variables). How can we get one more linear relationship between the roots?
The quantity \(A\) does not behave well when we swap the values of \(\alpha\) and \(\beta.\) Doing so leads us to the new value \[B = \beta + \omega\alpha + \omega^2\gamma.\]
If we can determine the value of \(B\) (in terms of our known quantities \(p, q, D,\) and \(A\)), then we would get our desired third linear equation relating the variables \(\alpha,\beta,\) and \(\gamma.\)
Fortunately, while \(A\) and \(B\) alone are not symmetric, the product \(AB\) is symmetric in \(\alpha,\beta,\gamma.\) Indeed, one can expand out \(AB\) to find \[AB = -\omega(\alpha\beta + \beta\gamma + \gamma\alpha) + \omega(\alpha^2 + \beta^2 + \gamma^2),\] which is manifestly symmetric.
Indeed, \(\alpha\beta + \beta\gamma+\gamma\alpha = p\) by Vieta's formulas, and \[\alpha^2 + \beta^2 + \gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta +\beta\gamma+\gamma\alpha) = 0 - 2p.\]
So, we now know the value of \(B\): it is just \(-3\omega p/A.\)
Now, to find our roots! We have the three linear equations \[\alpha + \beta + \gamma = 0,\] \[\alpha + \omega\beta + \omega^2\gamma = A,\] and \[\omega\alpha + \beta + \omega^2\gamma = B.\]
If we multiply the second equation by \(\omega^2,\) we get \[\alpha + \omega^2\beta + \omega\gamma = \omega^2 B.\] Using that \(1 + \omega + \omega^2 = 0,\) adding up our equations gives \[3\alpha + 0\beta + 0\gamma = 0 + A + \omega^2B.\] Thus \[\alpha = \frac{A + \omega^2B}{3}.\]
As swapping \(\alpha\) and \(\beta\) interchanges \(A\) and \(B,\) we deduce \[\beta = \frac{\omega^2A + B}{3},\] and therefore \[\gamma = \frac{\omega A + \omega B}{3}.\]
These are the three solutions to our cubic! Rewritten to eliminate \(B,\) we get \[\alpha = \frac{A}{3} - \frac{p}{A},\] \[\beta = \frac{\omega^2 A}{3} - \frac{\omega p}{A},\] \[\gamma = \frac{\omega A}{3} - \frac{\omega^2 p}{A}.\]
We remark that the system of equations we solved here may remind you of the discrete Fourier transform, which we discussed in Infinitely many primes end in a 7. And indeed, secretly the reason that this system of equations was so quick to solve was because of its relation to this discrete Fourier transform.
As our derivation of the cubic formula occupied many steps, let's use it on an explicit example now, just to make sure we understand what's going on. Let's solve the cubic equation \[x^3+6x^2+9x+3=0\] from the beginning of the article. By the change of variables \(x=y-2\) we converted it to the depressed cubic \[y^3 - 3y + 1 = 0.\] Earlier, we saw it had approximate solutions \(y \approx -1.88, y \approx 0.35, y \approx 1.53.\) Let's use our cubic formula to compute these roots exactly!
Here we have \(p = -3, q = 1.\) We start by needing to choose a square root of \[D^2 = -4p^3 - 27q^2 = -4 \cdot (-27) - 27 = 81.\] The number \(81\) has two square roots; let's go with the positive one, and choose \(D = 9\) for now.
We next need to choose a cube roots \[A^3 = -\frac{27}{2}q - \frac{3i\sqrt{3}}{2}D\]
Plugging in \(q = 1, D = 9,\) we find \[A^3 = \frac{-27 - 27i\sqrt{3}}{2} = 27\omega^2,\]
If we write \(\omega^2\) in polar form as \(\omega^2 = e^{4\pi i/3},\) we can deduce that one possible cube root of \(\omega\) is \[e^{4\pi i/9} = \cos\left(\frac{4\pi}{9}\right) + i\sin\left(\frac{4\pi}{9}\right).\]
So, we can choose \[A = 3e^{4\pi i/9} = 3\cos\left(\frac{4\pi}{9}\right) + 3i\sin\left(\frac{4\pi}{9}\right).\]
Plugging in these values into our cubic formula above, we get \[\alpha = e^{4\pi i/9} + e^{-4\pi i/9} = 2\cos\left(\frac{4\pi}{9}\right),\] \[\beta = e^{16\pi i/9} + e^{2\pi i/9} = 2\cos\left(\frac{2\pi}{9}\right),\] \[\gamma = e^{10\pi i/9} + e^{8\pi i/9} = 2\cos\left(\frac{8\pi}{9}\right).\] In fact, the cubic equation \(y^3 - 3y + 1 = 0\) is simply the triple-angle formula of cosine \[ \cos(3\theta)=4\cos^3(\theta)-3\cos(\theta). \]
With a calculator, we can check that \[2\cos(4\pi/9) \approx 0.35...,\] \[2\cos(2\pi/9) \approx 1.53...,\] \[2\cos(8\pi/9) \approx -1.88...,\] in agreement with the approximate solutions we found before! To go back to the equation \(x^3+6x^2+9x+3=0\) we recall we used the change of variables \(x=y-2\), so we just subtract 2 from each solution to get the solutions \[2\cos(4\pi/9)-2 \approx -1.65...,\] \[2\cos(2\pi/9)-2 \approx -0.47...,\] \[2\cos(8\pi/9)-2 \approx -3.88...,\]
We remark that there are, in general, two possible choices of \(D\) (the two different square roots) and three possible choices of \(A\) (the three different cube roots). This means there are \(2 \cdot 3 = 6\) possible choices we could have made; these choices correspond to the 6 different permutations of our roots. If we had chosen \(D = -9\) earlier, or if we had used a different cube root of \(27\omega^2,\) then we would have gotten the same roots, but we'd have assigned the labels \(\alpha,\beta,\) and \(\gamma\) differently.
We have now derived the cubic formula, and checked that it worked on a concrete cubic polynomial.
In our derivation, we used the fundamental theorem of symmetric polynomials: the fact that any symmetric polynomial in \(\alpha,\beta,\gamma\) can be rewritten as a polynomial in terms of the elementary symmetric sums \(\alpha+\beta+\gamma, \alpha\beta+\beta\gamma+\gamma\alpha,\) and \(\alpha\beta\gamma.\)
At some point in our derivation, we also had to rewrite the non-symmetric polynomial \(A^3\) in terms of \(p, q,\) and \(D.\) This is not quite the fundamental theorem of symmetric polynomials -- \(A^3\) is not symmetric, and \(D\) wasn't one of the elementary symmetric sums -- but it also isn't that far from it.
Galois theory is a vast generalization of the fundamental theorem of symmetric polynomials; just as the fundamental theorem of symmetric polynomials allows you to rewrite any symmetric polynomial in terms of simpler polynomials, Galois theory allows one to understand certain almost-symmetric polynomials in terms of more basic operations.
To develop Galois theory, one also needs to understand symmetries in a more tangible way. This leads to the concept of a group, a mathematical formalization of symmetry. A key insight of mathematics is that having the correct language to discuss an idea can often greatly simplify your work; for example, having the language of the polar form of a complex number made it much easier for us to compute the cube root of \(\omega^2\) earlier. The concept of a group has proven to be an incredibly useful language for describing symmetry, and having it is what gives Galois theory its power.