$p^3+q^3+r^3$, where $p$,$q$, and $r$ are the roots of the cubic function $x^3+4x^2-4x+1$. Working included.

Question

I am trying to utilise the expressions for Vieta's fomulae to solve expressions, just as an investigation. The question I gave myself is, if $p$, $q$, and $r$ are the roots of a cubic, what is $p^3+q^3+r^3$? The cubic is $x^3+4x^2-4x+1.$

I have done all the working out and come to the conclusion that the answer is $-115$. However, I am not 100% sure that my working out is correct, or if I used the formulae in the right way. If someone could check my working that I have attached as an image and point out any fundamental errors, that would be great. Thanks!

WORKING HERE

Answer 1

Your answer is correct. Here is an alternative approach.

You have $$p^2+q^2+r^2=(\underbrace{p+q+r}_{=-4})^2-2(\underbrace{pq+pr+qr}_{=-4})=24,$$ By $x^3=-4x^2+4x-1$, $$p^3+q^3+r^3=-4(p^2+q^2+r^2)+4(p+q+r)-(1+1+1)=-115.$$

Answer 2

Your solution is correct. You can also take a shortcut: $$\begin{align}p+q+r&=-4 \Rightarrow \\ p+q&=-4-r \Rightarrow \\ (p+q)^3&=(-4-r)^3 \Rightarrow \\ p^3+q^3+3pq(p+q)&=-64-48r-12r^2-r^3 \Rightarrow \\ p^3+q^3+r^3&=-3\left(-\frac1r\right)(-4-r)-64-48r-12r^2=\\ &=-12r^2-48r-67-\frac{12}{r}=\\ &=\frac{-12(r^3+4r^2-4r+1)-115r}{r}=\\ &=-115, \end{align}$$ because: $$\begin{align}p+q+r&=-4;\\ pq+qr+rp&=-4;\\ pqr&=-1;\\ r^3+4r^2-4r+1&=0.\end{align}$$