Question

If x=1-root(2),then find the value of (x-1/x)whole cube

Answer 1

Therefore 8 is the answer hope it helps

Answer 2

Answer:

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

Step-by-step explanation:

$Given \: x = 1-\sqrt{2}\:---(1)$

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

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

$Now,\\\left(x-\frac{1}{x}\right)^{3}\\=\left(1-\sqrt{2}-[-(1+\sqrt{2})]\right)^{3}\\=\left(1-\sqrt{2}+1+\sqrt{2}\right)^{3}\\=2^{3}\\=8$

Therefore,

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

•••♪