Question

If x = 2 + 2^1/3 + 2^2/3 , find the value of (x^3 + 6.x^2 + 6x). pls ⏩​

Answer 1

Explanation:

Given that x = 2 + 2^(2/3) + 2^(1/3)

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

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

use (a-b)^3= a^3 - b^3 - 3ab(a-b)

and (a+b)^3= a^3 + b^3 + 3ab(a+b) formulae.

or x^3 - 8 - 6x(x-2) = 2^2 + 2^1 + 3*[2^{(2/3)+(1/3)}[2^(2/3)+2^(1/3)

or x^3 - 8 - 6x^2 + 12x = 4 + 2 + 6(x-2)

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

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

Answer 2

Explanation:

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

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

$\bold{Cubing\;on\;both\;sides,\;We\;get\;:}$

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

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

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

$\implies x^3 - 6x^2 + 12x - 8 = 4 + 2 + 3(2^\frac{4}{3}) + 3(2^\frac{5}{3})$

$\implies x^3 - 6x^2 + 6x = 8 + 6 + 3(2^\frac{4}{3}) + 3(2^\frac{5}{3}) - 6x$

$\bold{But,\;We\;know\;that\;: x = 2 + 2^\frac{2}{3} + 2^\frac{1}{3}}$

$\implies x^3 - 6x^2 + 6x = 14 + 3(2^\frac{4}{3}) + 3(2^\frac{5}{3}) - 6(2 + 2^\frac{2}{3} + 2^\frac{1}{3})$

$\implies x^3 - 6x^2 + 6x = 14 + 3(2^\frac{4}{3}) + 3(2^\frac{5}{3}) - 12 - 6(2^\frac{2}{3}) - 6(2^\frac{1}{3})$

$\implies x^3 - 6x^2 + 6x = 2 + 3(2^\frac{4}{3}) + 3(2^\frac{5}{3}) - 3(2^(^\frac{2}{3}^+^1^)) - 3(2^(^\frac{1}{3}^+^1^))$

$\implies x^3 - 6x^2 + 6x = 2 + 3(2^\frac{4}{3}) + 3(2^\frac{5}{3}) - 3(2^\frac{4}{3}) - 3(2^\frac{5}{3})$

$\bold{\implies x^3 - 6x^2 + 6x = 2}$