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

If $\frac{\sqrt{3+x}+\sqrt{3-x}}{\sqrt{3+x}-\sqrt{3-x}}=2,$ then $x$ is equal to

The value of $1 + 1 + \frac{1}{4 \times 3} + \frac{1}{4 \times 3^{2}} + \frac{1}{4 \times 3^{3}}$ up to four places of decimals is

What will come in the place of the question mark $(?)$ in the following equation?
$2 \frac{1}{8} - 1 \frac{1}{16} = ? + 1 \frac{1}{32} - 1 \frac{9}{64}$

What is the value of $9 \frac{1}{3} + 19 \frac{2}{3} + 20 \frac{3}{4} + 19 \frac{1}{4}$?

If $5 \sqrt{5} \times 5^{3} \div 5^{-\frac{3}{2}} = 5^{a+2},$ then the value of $a$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module