if x=2+root 3, then find the value of (x-1/x)cube - Maths - Number Systems

Question

if x=2+root 3, then find the value of (x-1/x)cube

Vimala · Accepted Answer

Let $x = 2 + \sqrt{3}$

We need to find, ${(x - \frac{1}{x})}^{3}$ .

Since $x = 2 + \sqrt{3}$ , we have

$\frac{1}{x} = \frac{1}{2 + \sqrt{3}} = \frac{1}{2 + \sqrt{3}} \times \frac{2 - \sqrt{3}}{2 - \sqrt{3}} [m u l t i p l y a n d d i v i d e b y t h e c o n j u g a t e o f 2 + \sqrt{3}] = \frac{2 - \sqrt{3}}{2^{2} - (\sqrt{3})^{2}} [U s e t h e i d e n t i t y (a + b) (a - b) = a^{2} - b^{2}] = \frac{2 - \sqrt{3}}{4 - 3} = 2 - \sqrt{3}$

Thus,

${(x - \frac{1}{x})}^{3} = {(2 + \sqrt{3} - 2 + \sqrt{3})}^{3} = (2 \sqrt{3})^{3} = 8 \times 3 \times \sqrt{3} = 24 \sqrt{3}$

Thus, ${(x - \frac{1}{x})}^{3} = 24 \sqrt{3}$