Question 596131
<pre>
Notice that x²+x+1 one of the factors of x³-1 because

x³-1 = (x-1)(x²+x+1)

Then since x²+x+1 = 0, 

x³-1 = (x-1)(x²+x+1) = (x-1)(0) = 0

and therefore

 x³-1 = 0
   x³ = 1 
    x = {{{root(3,1)}}} 
    x = 1

Therefore  (x³+{{{1/x^3}}})³ = (1³+{{{1/1^3}}})³ = (1+{{{1/1}}})³ = (1+1)³ = 2³ = 8

Edwin</pre>