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

Correct option is B

Given:

​$$x = 2^{1/3} + 2^{2/3} + 2$$​​

Formula Used:

​$$(a + b)^3 = a^3 + b^3 + 3ab(a + b)$$​​

Solution:

​$$x = 2^{1/3} + 2^{2/3} + 2$$​​

​$$x - 2 = 2^{1/3} + 2^{2/3}$$​​

Cube both sides:

​$$(x - 2)^3 = \left(2^{1/3} + 2^{2/3}\right)^3$$​​

left-hand side:

​$$(x - 2)^3 = x^3 - 3x^2 \cdot 2 + 3x \cdot 4 - 8 = x^3 - 6x^2 + 12x - 8$$ 

Right - hand side ;

​$$\left(2^{1/3} + 2^{2/3}\right)^3 \\ \ \\ = 2^{(1/3)\times3} + 2^{(2/3)\times 3}+3 \times2^{1/3}\times2^{2/3}(2^{1/3} + 2^{2/3}) \\ \ \\ =2+4+6(x-2) \quad \quad [ 2^{1/3} + 2^{2/3} = x-2] \\ \ \\ =6 +6x-12 \\ \ \\= -6 + 6x$$​​

Now, Equating both sides:

​$$x^3 - 6x^2 + 12x - 8 = 6x - 6$$​​

​$$x^3 - 6x^2 + 12x - 8 - 6x + 6 = 0$$​​

​$$x^3 - 6x^2 + 6x - 2 = 0$$​​

​$$x^3 - 6x^2 + 6x = 2$$​​