Evaluating $x^3-18x^2+90x$, where $x= 6 + 6^{2/3}+ 6^{1/3}$ [duplicate]

Question

If $x= 6 + 6^{2/3}+ 6^{1/3}$, then find the value of $x^3-18x^2+90x$

My try:

I tried like directly substituting the value but it proved to be quite tedious and then I tried to take $6^{\frac{1}{3}}$ common but it didn't either.

Answer 1

Here is a solution using field theory. Perhaps not in the spirit of the question but fun anyway.

Let $\theta=6^{\frac{1}{3}}$ and consider $K=\mathbb Q(\theta)$, a cubic extension of $\mathbb Q$ having basis $\{1,\theta,\theta^2\}$.

With respect to this basis, the linear map $\mu: t \mapsto xt$ is given by this matrix: $$ X=\begin{pmatrix} 6 & 6 & 6 \\ 1 & 6 & 6 \\ 1 & 1 & 6 \end{pmatrix} $$ The characteristic polynomial of $X$ is $t^3 - 18 t^2 + 90 t - 150$ and so $x^3-18x^2+90x=150$.

The point here is that $x$, $\mu$, and $X$ satisfy exactly the same set of polynomials.

Answer 2

Here's my attempt at this:

Using the comment from @Vasili, we can plug $6+6^{\frac13}+6^{\frac23}$ in for $x$ right away to get$$(x-6)^3-18x+216=(6^{1/3}+6^{2/3})^3-6\cdot3\cdot(6+6^{1/3}+6^{2/3})+216\\42+216+18+108+18\sqrt[3]6+18\sqrt[3]6+18\sqrt[3]{36}\\=384+36\sqrt[3]6+18\sqrt[3]{36}$$which is the value that you need to solve this question.