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

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.

• I'm sure there must be better approach, but the algebra/arithmetic might go quicker if you rewrite the value of $x$ as $6^{2/3}(6^{-1/3} + 1 + 6^{1/3}).$ Still seems rather tediously pointless though (using my suggested rewrite), so that's surely not what is intended. Nov 20 at 20:58
• Recognize that $x^3-18x^2+90x=(x-6)^3-18x+216$ Nov 20 at 21:16
• Just factor it out using laws of exponents. Nov 20 at 22:47

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.

• – lhf
Nov 20 at 23:46

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$$(x-6)^3-18x+216=(6^{1/3}+6^{2/3})^3-6\cdot3\cdot(6+6^{1/3}+6^{2/3})+216\\42+216+18+108+18\sqrt6+18\sqrt6+18\sqrt{36}\\=384+36\sqrt6+18\sqrt{36}$$which is the value that you need to solve this question.

• No, correct answer is $150$
– Leox
Nov 20 at 22:47
• @Leox I'll check quickly to see where I went wrong Nov 20 at 23:01