Question

Suppose $a,b,c$ are the roots of the cubic $x^3-x^2-2=0$. Then the value of $a^3+b^3+c^3$ is _____

Answer 1

Answer: 7

Given $a+b+c=1$
$ab+bc+ca=0$
$bcc=2$
Now $(a+b+c{)}^2=1$
$a^2+b^2+c^2+2\sum ab=1$
$\therefore a^2+b^2+c^2=1$
Now, $a^3+b^3+c^3-3abc=(a+b+c)\left[\sum a^2-\sum ab\right]$
$=1(1-0)=1$
$a^3+b^3+c^3=1+3abc=1+3\times 2=7$