-1

Options are as follows:

  • $-6$
  • $0$
  • $1$
  • $2$
Marc van Leeuwen
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raghav rajput
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    Welcome to MSE. What have you tried? Where are you stuck? Is your calculation wrong? What do you exactly need help with? Questions showing no effort tend to be voted down, and will be closed – Andrei Jun 18 '19 at 11:10
  • Maybe this will help: https://en.wikipedia.org/wiki/Vieta%27s_formulas – Matti P. Jun 18 '19 at 11:11
  • Note that $x_i^9=(x_i^3)^3 = (1-x_i)^3$. Using tricks like this will allow you to transform the formula to the form where you can use Vieta's formulae. – Adam Latosiński Jun 18 '19 at 11:19
  • Multiple-choice is not an appropriate format for this site: what evidence do you have that the answer must be in the list? – Marc van Leeuwen Jun 18 '19 at 11:20

1 Answers1

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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!

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