Question

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

• A. $3$
• B. $6$
• C. $-2$
• D. $2$

Answer 1

The correct option is A $3$

$x=2+2^{1/3}+2^{2/3}x=2+2^{2/3}+2^{1/3}a=2r=2^{-1/3}n=3\therefore x=\frac{a(1-r^n)}{1-r}(sumofntermsofaG.P)x=\frac{2(1-2^{-1/3\ast 3})}{1-2^{-1/3}}x=\frac{2(1-2^{-1})}{1-2^{1/3}}x=\frac{1}{\frac{2^{1/3}-1}{2^{1/3}}}x=\frac{2^{1/3}}{2^{1/3}-1}{2}^{1/3}x-x=2^{1/3}{2}^{1/3}x-2^{1/3}=x{2}^{1/3}(x-1)=x2{(x-1)}^3=x^{3}2x^3-6x^2+6x-2=x^3{x}^3-6x^2+6x=2$