Mastering Non-linear Systems of Equations: Solving an Order 3 system

In this insightful video, I'll show you how we can tackle a challenging system of equations using an unconventional approach.

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[XJWill]

You essentially derived the Newton-Girard identities for a 3 variable system. It is easy enough to derive them in this case, but for similar but more complicated problems, it is much easier to just use the Newton identities.
For reference, here are the definitions and identities for a 3 variable system:
Elementary symmetric polynomials:
e0 = 1
e1 = x + y + z
e2 = x*y + x*z + y*z
e3 = x*y*z
e4, e5, e6... = 0 (since we are looking at a 3 variable system)
Symmetric power polynomials:
p1 = x^1 + y^1 + z^1
p2 = x^2 + y^2 + z^2
p3 = x^3 + y^3 + z^3
pn = x^n + y^n + z^n
Newton-Girard identities:
1*e1 = e0*p1
2*e2 = e1*p1 - e0*p2
3*e3 = e2*p1 - e1*p2 + e0*p3
4*e4 = e3*p1 - e2*p2 + e1*p3 - e0*p4
....
We already know most of the values: e1 = p1, p2, p3 and we only need to solve for e2 and e3:
2*e2 = (2*a)*(2*a) - 1*6*a^2 => e2 = -a^2
3*e3 = (-a^2)*(2*a) - (2*a)*(6*a^2) + 1*(8*a^3) => e3 = -2*a^3
Now we use Vieta's formula to construct the cubic equation
t^3 - e1*t^2 + e2*t - e3 = 0
t^3 - 2*a*t^2 - a^2*t + 2*a^3 = 0
Which has roots:
t1 = a
t2 = -a
t3 = 2*a

Of course, that seems a lot of extra work to do for this simple problem. But like I said, this method easily extends to more complicated problems where the direct method would be more work than using the Newton-Girard identities.