Math, asked by greeshma13karumanchi, 1 month ago

If x= 2 + 2^2/3 + 2^1/3 , then find the value of x^3 - 6x² + 6x.​

Answers

Answered by Salmonpanna2022
3

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}

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