[Math] If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$.

Question

If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, find the value of $p^3+q^3+r^3$.

Here's what I have got,

By Vieta's rule

$p+q+r=1\text{. ………..}(1)$

$pq+qr+pr=1\text{. ………..}(2)$

$pqr=2\text{. ………..}(3)$

Squaring $(1)$,

$p^2+q^2+r^2+2(pq+qr+pr)=1\text{. ………..}(4)$

From $(2)$,

$p^2+q^2+r^2=-1\text{. ………..}(5)$

Putting the roots and adding these equations,

$p^3-p^2+p-2=0$

$q^3-q^2+q-2=0$

$r^3-r^2+r-2=0$

We get,

$(p^3+q^3+r^3)-(p^2+q^2+r^2)+(p+q+r)-6=0$

Putting the values,

$(p^3+q^3+r^3)-(-1)+1-6=0$

$(p^3+q^3+r^3)=4$

Am I doing something wrong in my solution?

Because the answer given is -5.

Any help would be appreciated.

Answer 1

You can use the strict $p^3+q^3+r^3-3pqr=(p+q+r)(p^2+q^2+r^2-pq-qr-rp)$. It's easy to compute $p^2+q^2+r^2$ and use Viete's rule.