Question

If x=2=2^2/3+2^1/3,then find the value of x^3-6x^2+6x

Answer 1

Answer:

pubg khelo or mast raho

Step-by-step explanation:

pls

Answer 2

Step-by-step explanation:

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

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

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

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

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

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

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

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

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

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

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

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

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

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