An algebra problem by Swastik Sahoo

x=2+2^1/3+2^2/3

bringing 2 on L.H.S and cubing on b.s

(x-2)^3=(2^1/3+2^2/3)^3

expanding.......... x^3-8-6x(x-2)=2+4+6(2^1/3+2^2/3)

but 2^1/3+2^2/3=x-2.........from the first step

therefore R.H.S=2+4+6(x-2)=6+6x-12

hence x^3-6x^2+6x=6+6x-12

finally x^3-6x^2+6x-2=0

Therefore answer is 0

$x = 2+2^\dfrac{1}{3}+2^\dfrac{2}{3}$ ,

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

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

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

cubing on both the sides

$x^3 = 2(x^3-3x^2+3x-1)$

$x^3-6x^2+6x-2 = 0$