Q13/001) If x^2+x+1=0,then find the value of [x^3+1/x^3]^3

x^2 + x + 1 = 0
so x^2 + 1 = - x
(x + 1/x) = -1


cube both sides

(x^3 + 1/x^3) + 3(x+ 1/x) = - 1
or (x^3 + 1/x^3) - 3 = - 1
or (x^3 + 1/x^3) = 2 hence
(x^3 + 1/x^3)^3 = 8