Question

If $x=3+3^{\frac{1}{3}}+3^{\frac{2}{3}}$, then find the value of $x^3-9x^2+18x$.

Answer 1

$x=3+3^{1/3}+3^{2/3}$

$\Rightarrow x-3=3^{1/3}+3^{2/3}$

Cubing both side we get

$x^3-3^3-3.x.3\left(x-3\right)=3^{3/3}+3^{6/3}+{3.3}^{1/3}{.3}^{2/3}\left(3^{1/3}+3^{2/3}\right)$

$\Rightarrow x^3-27-9x^2+27x=3+9+9\left(x-3\right)$

$\Rightarrow x^3-9x^2+27x-27=12+9x-27$

$\Rightarrow x^3-9x^2+18x-12=0$

$\Rightarrow x^3-9x^2+18x=12$