Question

if x is equals to 1 minus under root 2 find the value of x minus one upon X whole cube

Answer 1

Answer:  8

Step-by-step explanation:

given that

x= 1-√2

so

$\frac{1}{x}=\frac{1}{1-\sqrt{2}}\\\\=\frac{1}{1-\sqrt{2}}*\frac{1+\sqrt{2}}{1+\sqrt{2}}\\\\=\frac{1+\sqrt{2}}{1-2}\\\\-1-\sqrt{2}$

so the value of

$(x-\frac{1}{x})^{3}\\\\=(1-\sqrt{2}+1+\sqrt{2})^{3}\\\\=(2)^{3}\\\\=8$

Answer 2

Given:

$x=1-\sqrt 2$

To find:

Value of

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

Solution:

$\frac{1}{x}=\frac{1}{1-\sqrt{2}}$

By rationalization

$\frac{1}{x}=\frac{1+\sqrt{2}}{(1-\sqrt{2})(1+\sqrt{2})}$

$\frac{1}{x}=\frac{1+\sqrt{2}}{(1)^2-(\sqrt{2})^2}$

Using identity:

$(a+b)(a-b)=a^2-b^2$

$\frac{1}{x}=\frac{1+\sqrt{2}}{1-2}=-1-\sqrt{2}$

Now, substitute the values

$(x-\frac{1}{x})^3=(1-\sqrt{2}+1+\sqrt{2})^3$

$=(2)^3$

$=8$

Hence, the value of

$(x-\frac{1}{x})^3=8$