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.

Answer 2

$$(x^3-2)^3=(x^2-x)^3$$

$$(x^3)^3-8-3(x^3)^22+3(x^3)2^2=(x^3)^2-(x^3)-3x^2\cdot x(x^3-2)$$

Replace $x^3=y$ to find

$$y^3-8-3y^2\cdot2+3y\cdot2^2=y^2-y-3y(y-2)$$

$$\iff y^3-y^2(6+1-3)+\cdots=0$$ whose roots are $ p^3,q^3,r^3$

$$\implies p^3+q^3+r^3=\dfrac{6+1-3}1$$

Answer 3

$$p³+q³+r³ = (p+q+r)³-3(p+q)(q+r)(r+p)$$ $$ = (1)³- 3(1-r)(1-p)(1-q)$$ $$ = 1-3(1-1+1-2) =4$$ Here, 1-1+1-2= the sum of coefficients of the given equation