Question

if x=1-√2, find the value of (x-1/x)^3
please tell me the answer with proper explanation​

Answer 1

$\large\underline{\text{Question}}$

If $x=1-\sqrt{2}$, find the value of $\left(x-\dfrac{1}{x}\right)^{3}$.

$\large\underline{\text{Note}}$

There are two methods to solve this problem.

$\large\underline{\text{Solution 1. Rationalization}}$

We know that

$\red{\bigstar}x=1-\sqrt{2}$

$\implies\dfrac{1}{x}=\dfrac{1}{1-\sqrt{2}}$

$\implies\dfrac{1}{x}=\dfrac{1}{1-\sqrt{2}}\times\dfrac{1+\sqrt{2}}{1+\sqrt{2}}$

$\implies\dfrac{1}{x}=\dfrac{1+\sqrt{2}}{1-2}$

$\implies\dfrac{1}{x}=-1-\sqrt{2}$

(by adding $x$ and $\dfrac{1}{x}$)

$\implies x-\dfrac{1}{x}=2$

$\red{\bigstar}\boxed{\left(x-\dfrac{1}{x}\right)^{3}=8}$

$\large\underline{\text{Solution 2. Quadratic equation}}$

We can derive a quadratic equation from squaring both sides. Then, we can divide by $x$ to find the required value.

We know that

$\red{\bigstar}x=1-\sqrt{2}$

(isolating the irrational term)

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

(by squaring both sides)

$\implies (x-1)^{2}=2$

$\implies x^{2}-2x-1=0$

(divide both sides by $x\neq0$)

$\implies x-2-\dfrac{1}{x}=0$

$\implies x-\dfrac{1}{x}=2$

And therefore

$\red{\bigstar}\boxed{\left(x-\dfrac{1}{x}\right)^{3}=8}$

Answer 2

Thank you for the points.

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