Q.If \(x = 7 + 4\sqrt{3}\), then find the value of \(\cfrac{x^3}{x^6 + 7x^3 + 1}\). (a) \(\cfrac{1}{2737}\) (b) \(\cfrac{1}{2730}\) (c) \(\cfrac{1}{2710}\) (d) \(\cfrac{1}{2709}\)
Answer: D
Let \(x = 7 + 4\sqrt{3}\) \(\therefore \cfrac{1}{x} = \cfrac{1}{7 + 4\sqrt{3}}\) \(= \cfrac{(7 - 4\sqrt{3})}{(7 + 4\sqrt{3})(7 - 4\sqrt{3})}\) \(= \cfrac{(7 - 4\sqrt{3})}{49 - 48} = 7 - 4\sqrt{3}\) \(\therefore x + \cfrac{1}{x} = 7 + 4\sqrt{3} + 7 - 4\sqrt{3} = 14\) Now, \(\cfrac{x^3}{x^6 + 7x^3 + 1}\) \(= \cfrac{x^3}{x^3(x^3 + 7 + \cfrac{1}{x^3})}\) \(= \cfrac{1}{x^3 + \cfrac{1}{x^3} + 7}\) \(= \cfrac{1}{(x + \cfrac{1}{x})^3 - 3x \cdot \cfrac{1}{x} \cdot (x + \cfrac{1}{x}) + 7}\) \(= \cfrac{1}{14^3 - 3 \cdot 14 + 7}\) \(= \cfrac{1}{2744 - 42 + 7}\) \(= \cfrac{1}{2709}\)
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