Question

If $x^2+x+1=0$ then what is the value of ${(x^3+\frac{1}{x^3})}^3$ ?

Answer 1

Consider $x^2+x+1=0$ and multiply both sides by $(x-1)$

multiply with (x-1) on both sides as shown below:

$(x-1)(x^2+x+1)=0\times (x-1)\Rightarrow (x-1)(x^2+(x\times 1)+1^2)=0\Rightarrow x^3-1^3=0(\because a^3-b^3=(a-b\left)\right(a^2+ab+b^2)\Rightarrow x^3-1=0\Rightarrow x^3=1$

Now let us find ${(x^3+\frac{1}{x^3})}^3$ as follows:

${(x^3+\frac{1}{x^3})}^3={(1+\frac{1}{1})}^3={(1+1)}^3=2^3=8$

Hence, ${(x^3+\frac{1}{x^3})}^3=8$