The sum of the 9th power of the roots of the equation $x^3+x-1=0$ are:

Question

Options are as follows:

• $-6$
• $0$
• $1$
• $2$

Answer 1

Let the roots be $a,b$ and $c$. Since all of them satisfies the equation $x^3+x-1=0$, \begin{align} a^9+b^9+c^9 &= (1-a)^3+(1-b)^3+(1-c)^3\\ &= -(a^3+b^3+c^3)+3(a^2+b^2+c^2)-3(a+b+c)+3\\ &=-[(1-a)+(1-b)+(1-c)]+3[(a+b+c)^2-2(ab+bc+ca)]-3(a+b+c)+3\\ &=3(a+b+c)^2-6(ab+bc+ca)-2(a+b+c). \end{align} Further, Vieta's formulas gives $a+b+c$ and $ab+bc+ca$, which can be plugged into the above expression to get the final answer!