Question

If $x=2+2^{\frac{2}{3}}+2^{\frac{1}{3}}$, then the value of $x^3-6x^2+6x$, is

• A. $3$
• B. $2$
• C. $1$
• D. $-2$

Answer 1

Answer: (B)

$2$

We know that ${(a-b)}^3=a^3-3a^{2}b+3a{b}^2-b^3$
Here we have $x=2+2^{\frac{2}{3}}+2^{\frac{1}{3}}$
$\Rightarrow$ $x-2=2^{\frac{2}{3}}+2^{\frac{1}{3}}$
$\Rightarrow$ ${(x-2)}^3={(2^{\frac{2}{3}}+2^{\frac{1}{3}})}^3$
$\Rightarrow$ $x^3-3{\left(x\right)}^2\left(2\right)+3\left(x\right){\left(2\right)}^2-2^3={\left(2^{\frac{2}{3}}\right)}^3+3{\left(2^{\frac{2}{3}}\right)}^2\left(2^{\frac{1}{3}}\right)+3\left(2^{\frac{2}{3}}\right){\left(2^{\frac{1}{3}}\right)}^2+{\left(2^{\frac{1}{3}}\right)}^3$
$\Rightarrow$ $x^3-6x^2+12x-8=4+3\left(2^{\frac{5}{3}}\right)+3\left(2^{\frac{4}{3}}\right)+2$
$\Rightarrow$ $x^3-6x^2+12x=14+3\left(2^{\frac{5}{3}}\right)+3\left(2^{\frac{4}{3}}\right)$
$\Rightarrow$ $x^3-6x^2+6x=14+3\left(2^{\frac{5}{3}}\right)+3\left(2^{\frac{4}{3}}\right)-6x$
$\Rightarrow$ $x^3-6x^2+6x=14+3\left(2^{\frac{5}{3}}\right)+3\left(2^{\frac{4}{3}}\right)-6(2+2^{\frac{2}{3}}+2^{\frac{1}{3}})$
$\Rightarrow$ $x^3-6x^2+6x=2+3\left(2^{\frac{5}{3}}\right)-6\left(2^{\frac{2}{3}}\right)+3\left(2^{\frac{4}{3}}\right)-6\left(2^{\frac{1}{3}}\right)$
$\Rightarrow$ $x^3-6x^2+6x=2+3\left(2^{\frac{5}{3}}\right)-\left(3\right)\left(2\right)\left(2^{\frac{2}{3}}\right)+3\left(2^{\frac{4}{3}}\right)-\left(3\right)\left(2\right)\left(2^{\frac{1}{3}}\right)$
$\Rightarrow$ $x^3-6x^2+6x=2+3\left(2^{\frac{5}{3}}\right)-\left(3\right)\left(2^{\frac{5}{3}}\right)+3\left(2^{\frac{4}{3}}\right)-\left(3\right)\left(2^{\frac{4}{3}}\right)$
$\Rightarrow$ $x^3-6x^2+6x=2$
Hence, (B) is the correct answer.