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