Question:
\[ x – \frac{1}{x} = 3 + 2\sqrt{2} \]
Find: \[ x^3 – \frac{1}{x^3} \]
Solution:
Using identity:
\[ \left(a-b\right)^3 = a^3-b^3-3ab(a-b) \]
\[ \left(x-\frac{1}{x}\right)^3 = x^3-\frac{1}{x^3} – 3\left(x\right)\left(\frac{1}{x}\right) \left(x-\frac{1}{x}\right) \]
Since, \[ x\cdot \frac{1}{x}=1 \]
\[ \left(x-\frac{1}{x}\right)^3 = x^3-\frac{1}{x^3} – 3\left(x-\frac{1}{x}\right) \]
\[ x^3-\frac{1}{x^3} = \left(x-\frac{1}{x}\right)^3 + 3\left(x-\frac{1}{x}\right) \]
Substituting \[ x-\frac{1}{x}=3+2\sqrt{2} \]
\[ x^3-\frac{1}{x^3} = (3+2\sqrt{2})^3 + 3(3+2\sqrt{2}) \]
\[ = (17+12\sqrt{2})(3+2\sqrt{2}) + 9+6\sqrt{2} \]
\[ = 51+34\sqrt{2}+36\sqrt{2}+48 + 9+6\sqrt{2} \]
\[ = 108+76\sqrt{2} + 9+6\sqrt{2} \]
\[ = 117+82\sqrt{2} \]