$p^3+q^3+r^3$, 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

There's no image. But in any case, you should use MathJax rather than an image. — Robert Israel, Nov 29 '18 at 03:57
I have updated the post. Also, there are many lines of working and I'm new to Maths Stack Exchange so I feel that in this moment in time neat, written working is the most appropriate way to go :) — Chem_Student, Nov 29 '18 at 04:00
See https://math.stackexchange.com/questions/3015661/system-of-3-equations-problem — lab bhattacharjee, Nov 29 '18 at 04:03

score 1 · Accepted Answer · answered Nov 29 '18 at 04:05

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 $x^3=-4x^2+4x-1$, $$p^3+q^3+r^3=-4(p^2+q^2+r^2)+4(p+q+r)-(1+1+1)=-115.$$

answered Nov 29 '18 at 04:05

Tianlalu

5,177

Thank you for the quicker method, however I don't really understand if you used a formula for the first line, or if it's derived from elsewhere? – Chem_Student Nov 29 '18 at 04:11
Expand $(p+q+r)^2=p^2+q^2+r^2+2pq+2pr+2qr$. – Tianlalu Nov 29 '18 at 04:13
Ok thank you so much! – Chem_Student Nov 29 '18 at 04:15
Sorry, also, did you subsitute x=p+q+r, because I don't understand how $x^3=p^3+q^3+r^3$ – Chem_Student Nov 29 '18 at 04:20
Substitute $p,q,r$ separately, we get 3 equalities. Then add them together. – Tianlalu Nov 29 '18 at 04:22
$\begin{cases}p^3=-4p^2+4p-1\q^3=-4q^2+4q-1\r^3=-4r^2+4r-1\end{cases}$ – Tianlalu Nov 29 '18 at 04:25

score 0 · Answer 2 · answered Nov 29 '18 at 06:57

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: $$\begin{align}p+q+r&=-4;\\ pq+qr+rp&=-4;\\ pqr&=-1;\\ r^3+4r^2-4r+1&=0.\end{align}$$

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

2 Answers2