Cubic Root verification

Question

I'm working on a cubic root question and goes like this...

Let, r,s, and t be the roots of the equation $x^3+ax^2+bx+c = 0$. Re-write the expressions in terms of a,b, and c.

I'm given $r^2+s^2+t^2$. I know that, given 3 roots, I can factor the cubic equation as $(x-r)(x-s)(x-t)$, but don't know what to replace $x$ by in order to get $r^2+s^2+t^2$.

Answer 1

$$(r+s+t)^2=?$$

Now by Veita's formula $$r+s+t=-\dfrac a1$$ and $$rs+st+tr=\dfrac b1$$

Answer 2

We haven't studied Vieta's formula in my class just yet.

Pretending to not know Vieta's formulas, note that if:

$$P(x)=x^3+ax^2+bx+c = (x-r)(x-s)(x-t)$$

then:

$$P(-x)=-x^3+ax^2-bx+c = -(x+r)(x+s)(x+t)$$

It follows that:

$$ \begin{align} P(x)P(-x) & = (x^3+ax^2+bx+c)(-x^3+ax^2-bx+c) \\ & = (ax^2+c)^2-(x^3+bx)^2 \\ & = -x^6 +x^4(a^2-2b)+\cdots \end{align} $$

and also:

$$ \begin{align} P(x)P(-x) & = -(x^2-r^2)(x^2-s^2)(x^2-t^2) \\ & =-x^6+x^4(r^2+s^2+t^2)+\cdots \end{align} $$

Now equate the coefficient of $x^4$ between the two expressions.