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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

$\sqrt{? \times 7} \times 18 = 84$

If $a^{2}+b^{2}=45$ and $ab=18$,find $\frac{1}{a}+\frac{1}{b}$.

If $x=2$,then find the value of $x^{3}+27x^{2}+243x+631$.

$\frac{1}{\left(2 \frac{1}{3}\right)}+\frac{1}{\left(1 \frac{3}{4}\right)}$ is equal to

If $(2+\sqrt{3}) a=(2-\sqrt{3}) b=1,$ then the value of $\frac{1}{a}+\frac{1}{b}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module