The value of (3+2 sqrt{2})^{-3}+(3-2 sqrt{2}) | vedclass

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(A) Let $x = 3+2 \sqrt{2}$ and $y = 3-2 \sqrt{2}$.Note that $xy = (3+2 \sqrt{2})(3-2 \sqrt{2}) = 3^2 - (2 \sqrt{2})^2 = 9 - 8 = 1$.We need to find $x^

Vedclass · Accepted Answer

$198$

Vedclass · Answer

(A) Let $x = 3+2 \sqrt{2}$ and $y = 3-2 \sqrt{2}$.Note that $xy = (3+2 \sqrt{2})(3-2 \sqrt{2}) = 3^2 - (2 \sqrt{2})^2 = 9 - 8 = 1$.We need to find $x^{-3} + y^{-3} = \frac{1}{x^3} + \frac{1}{y^3} = \frac{x^3 + y^3}{(xy)^3}$.Since $xy = 1$,this simplifies to $x^3 + y^3$.Using the identity $x^3 + y^3 = (x+y)^3 - 3xy(x+y)$:$x+y = (3+2 \sqrt{2}) + (3-2 \sqrt{2}) = 6$.$x^3 + y^3 = (6)^3 - 3(1)(6) = 216 - 18 = 198$.

The value of $(3+2 \sqrt{2})^{-3}+(3-2 \sqrt{2})^{-3}$ is

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