Question 52

If $$x = \frac{3 + \sqrt{6}}{5\sqrt{3} - 2\sqrt{12} - \sqrt{32} + \sqrt{50}}$$, then $$\frac{x^4 - 1}{x^4 + 1}$$ =

Solution

$$5\sqrt{3\ }-2\sqrt{12\ }-\sqrt{32\ }+\sqrt{50\ }=5\sqrt{3\ }-4\sqrt{3\ }-4\sqrt{2\ }+5\sqrt{2\ }=\sqrt{3\ }+\sqrt{2\ }.$$

$$3+\sqrt{6\ }=\sqrt{3\ }\left(\sqrt{3\ }+\sqrt{2\ }\right).$$

So,$$x=\sqrt{3\ }.$$

So,$$\ \frac{\ x^4-1}{x^4+1}=\frac{9-1\ }{9+1}=\frac{8\ }{10}=\frac{4\ }{5}.$$

C is correct choice.

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