Question

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

• A. $3$
• B. $2$
• C. $1$
• D. None of these

Answer 1

Answer: (B)

$2$

$x=2+2^{\frac{2}{3}}+2^{\frac{1}{3}}$

$x-2=2^{\frac{2}{3}}+2^{\frac{1}{3}}$ ........ (1)

Cubing both sides,

$(x-2{)}^3={(2^{\frac{2}{3}}+2^{\frac{1}{3}})}^3$

$x^3-2^3-6x(x-2)={\left(2^{\frac{2}{3}}\right)}^3+{\left(2^{\frac{1}{3}}\right)}^3+3\times 2^{\frac{2}{3}}\times 2^{\frac{1}{3}}(2^{\frac{2}{3}}+2^{\frac{1}{3}})$

$x^3-8-6x^2+12x=4+2+6(x-2)$ ( from (1) $x-2=2^{\frac{2}{3}}+2^{\frac{1}{3}}$ )

$x^3-6x^2+12x-8=6+6x-12$

$x^3-6x^2+12x-6x=6-12+8$

$x^3-6x^2+6x=2$