Question

If $x=2+2^{1/3}+2^{2/3}$ then prove that $x^3-6x^2+6x=$

• A. $0$
• B. $2$
• C. $1$
• D. $6$

Answer 1

The correct option is B $2$

We have,

$x=2+2^{1/3}+2^{2/3}$

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

Cube above equation both sides

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

$\Rightarrow x^3-8-6x(x-2)=6+6(x-2)$, using (1)

$\Rightarrow x^3-6x^2+6x=2$, simplify

Note: $(a+b{)}^3=a^3+b^3+3ab(a+b)$