If $x = 3^{1/3} + 3^{2/3} + 3$, find the value of $x^3 - 9x^2 + 18x - 12$

Question

If $x = 3^{1/3} + 3^{2/3} + 3$, find the value of $$x^3 - 9x^2 + 18x - 12.$$

This is not a homework problem. I'm not even a student. I'm going through an old textbook. I know this is a simple problem. Can't seem to crack it though.

Answer 1

It makes sense to let $y = x - 3 = 3^{1 \over 3} + 3^{2 \over 3}$. You are to compute $$(y + 3)^3 -9(y + 3)^2 + 18(y + 3) - 12$$ This works out to $$y^3 - 9y - 12$$ Note that $y^3 = (3^{1 \over 3})^3(1 + 3^{1 \over 3})^3 = 3(1 + 3*3^{1 \over 3} + 3*3^{2 \over 3} + 3) = 3(4 + 3y) = 12 + 9y$. Thus $y^3 - 9y - 12 = 0$.

Answer 2

Suggestion: work directly by "foiling out" the products. While doing that use that two terms with the same base, raised to a power, mean add the exponents. Everything will then boil down to integers, or integers times $3^{1/3},$ or integers times $3^{2/3}.$ What I mean by foiling out is that, e.g., $x^2=(3^{1/3}+3^{2/3}+3)\cdot (3^{1/3}+3^{2/3}+3)$ for which each of the three terms in the first are to be multiplied by each of them in the second. Then simplify the resulting products by adding exponents as noted.

Answer 3

Put $t=\sqrt[3]{3}$, so we have $t^3=3.$ Then $$ x=t+t^2+t^3=t(1+t+t^2)=t \frac{t^3-1}{t-1}=\frac{2t}{t-1}. $$ Now substitute this $x$ to ${x}^{3}-9\,{x}^{2}+18\,x-12$: $$ {x}^{3}-9\,{x}^{2}+18\,x-12=8\,{\frac {{t}^{3}}{ \left( t-1 \right) ^{3}}}-36\,{\frac {{t}^{2}}{ \left( t-1 \right) ^{2}}}+36\,{\frac {t}{t-1}}-12=\ldots=-4\,{\frac {{t}^{3}-3}{ \left( t-1 \right) ^{3}}}=0. $$

Answer 4

Let $\,a = \sqrt[3]3.\, $ $\, (\!\overbrace{x\!-\!3}^{\Large a^2+a}\!)^2 = \overbrace{a^4}^{\Large 3a}\!+\overbrace{2a^3}^{\Large 6}+a^2\,$

Therefore $\ \color{#0a0}{x^2\!-6x\!+\!3\, =\, a^2\!+\!3a}.\ $ By very simple arithmetic we have:

$\begin{align} f(x)\, & =\ (x-3)\,(\color{#0a0}{x^2\!-6x+3})\, -\, 3(x+1)\ \ {\rm\ via\ \ Division\ Algorithm}\\ & = (a^2\!+\!a)\,\color{#0a0}{(a^2\!+\!3a)}\, -\, 3(a^2+a+4)\\ &= \underbrace{a^4}_{\Large 3a}\!\!+\underbrace{4a^3}_{\Large\! 4\,\cdot\, 3}\!+3a^2\, -\, 3(a^2+a+4)\, =\, 0 \end{align}$