Question

if x is equal to 2 + root 3 find x minus one by X whole cube​

Answer 1

Answer:

Step-by-step explanation:It is given that x=2+$\sqrt3$

x minus one by X = 2-$\sqrt{3}$

x minus one by X whole cube​=26-15$\sqrt{3}$

Answer 2

The solution is done below for you :

Given : $\mathrm{x=2+\sqrt{3}}$

To find : $\mathrm{(x-\frac{1}{x})^{3}}$ = ?

Solution :

Here, $\mathrm{x=2+\sqrt{3}}$

Then, $\mathrm{\frac{1}{x}=\frac{1}{2+\sqrt{3}}}$

We multiply $\mathrm{(2-\sqrt{3})}$ to both the numerator and the denominator to rationalise the denominator:

$\quad\quad \mathrm{=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}}$

$\quad\quad \mathrm{=\frac{2-\sqrt{3}}{4-3}}$

$\quad\quad \mathrm{=2-\sqrt{3}}$

Then $\mathrm{x-\frac{1}{x}}$

$\quad \mathrm{=(2+\sqrt{3})-(2-\sqrt{3})}$

$\quad \mathrm{=2+\sqrt{3}-2+\sqrt{3}}$

$\quad \mathrm{=2\sqrt{3}}$

$\implies \mathrm{x-\frac{1}{x}=2\sqrt{3}}$

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

$\quad\quad\quad\quad\quad \mathrm{=2^{3}\times (\sqrt{3})^{3}}$

$\quad\quad\quad\quad\quad \mathrm{=8\times 3\sqrt{3}}$

$\quad\quad\quad\quad\quad \mathrm{=24\sqrt{3}}$

$\therefore \boxed{\mathrm{\quad(x-\frac{1}{x})^{3}=24\sqrt{3}\quad}}$