Question

Do it on paper and explain it
If x=3+2√2 find the value of (x-1/x)^3

Answer 1

Answer:

128√2

Step-by-step explanation:

${\underline{{\text{Given}}}}$

$x = 3 + 2 \sqrt{2}$

${\underline{{\text{To Find:}}}}$

$(x - \dfrac{1}{x} ) {}^{3}$

${\underline{{\text{Solution:}}}}$

First We should find the value of $\frac{1}{x}$

$x = 3 + 2 \sqrt{2} \\ \\ \text{\text{\text{Or,}}} \: \: \frac{1}{x} = \frac{1}{3 + 2 \sqrt{2} } \\ \\ \text{ [Rationalizing the denominator }]\\ \\ = \frac{(3 - 2\sqrt{2}) }{(3 + 2 \sqrt{2})(3 - 2 \sqrt{2}) } \\ \\ = \frac{3 - 2 \sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2}) }^{2} } \: \: \text{[using (a+b)(a-b) = } {a}^{2} - {b}^{2} ]\\ \\ = \frac{3 - 2 \sqrt{2} }{9 - 8} \\ \\ = \frac{3 - 2 \sqrt{2} }{1} \\ \\ \therefore \: \frac{1}{x} = 3 - 2 \sqrt{2}$

Now,

$(x - \frac{1}{x} ) {}^{3} \\ \\ \{ (3 + 2 \sqrt{2} ) - (3 - 2 \sqrt{2} )\} {}^{3} \\ \\ (3 + 2 \sqrt{2} - 3 + 2 \sqrt{2} ) {}^{3} \\ \\ {(4 \sqrt{2} )}^{3} \\ \\ 64 \times 2 \sqrt{2} \\ \\ 128 \sqrt{2} \\ \\ \therefore\text{ The required answer is 128 \sqrt{2} }$

${\underline{{\text{ Some Important Formula Of Algebra }}}}$

${(x + y)}^{2}={x}^{2}+{y}^{2}+2xy\\ \\{(x - y)}^{2}={x}^{2}+{y}^{2}-2xy\\ \\{(x+y)}^{2}= (x - y) {}^{2}+4xy\\ \\{(x-y)}^{2}=(x+y){}^{2}-4xy\\ \\ (x + y)^{2}+(x-y)^{2}=2( {x}^{2}+{y}^{2} )\\ \\(x+y)^{2}- (x-y) {}^{2}=4xy\\ \\ {(x + y)}^{3}={x}^{3}+{y}^{3}+ 3xy(x + y) \\ \\(x - y)^{3}={x}^{3}-{y}^{3}- 3xy(x - y)$