Question

if x = 3+2√2 then find the value of (x-1/x )^3 ​

Answer 1

Answer:

$\large (x-\dfrac{1}{x})^3=128\sqrt2$

Step-by-step explanation:

Given :

$\large x=3+2\sqrt2$

Now taking reciprocal of both side

$\large \dfrac{1}{x}=\dfrac{1}{3+2\sqrt2}$

Now rationalize the denominator

$\large \dfrac{1}{x}=\dfrac{1}{3+2\sqrt2}\times\dfrac{3-2\sqrt2}{3-2\sqrt2} \\\\\\\large \dfrac{1}{x}=\dfrac{3-2\sqrt2}{9-8}\\\\\\\large \dfrac{1}{x}=\dfrac{3-2\sqrt2}{1}\\\\\\\large \dfrac{1}{x}= 3-2\sqrt2$

On subtracting we get

$\large x-\dfrac{1}{x}= 3+2\sqrt2-(3-2\sqrt2)\\\\\\\large x-\dfrac{1}{x}= 3-3+2\sqrt2+2\sqrt2\\\\\\\large x-\dfrac{1}{x}= 4\sqrt2$

We have to find  $\large (x-\dfrac{1}{x})^3$

$\large \text{Rewrite (x-\dfrac{1}{x})^3 as (x-\dfrac{1}{x})(x-\dfrac{1}{x})(x-\dfrac{1}{x}) }$

Put value here we get.

$\large (x-\dfrac{1}{x})^3=4\sqrt2\times4\sqrt2\times4\sqrt2\\\\\\\large (x-\dfrac{1}{x})^3=4\times4\times4\times\sqrt2\times\sqrt2\times\sqrt2\\\\\\(x-\dfrac{1}{x})^3=64\times2\sqrt2\\\\\\(x-\dfrac{1}{x})^3=128\sqrt2$

Thus we get answer.

Answer 2

we have

X = 3+2√2

✨if yr question

is

[(X+1)/X]^3

then answer is -2-2√2

if yr question is [x-(1/X)] ^3

is 128√2

.•♫•♬•[ I hope it help you ❤️ ].•♫•♬•

Attachment!!!