If $$x + \frac{1}{x} = 3\sqrt{2}$$, then what is the value of $$x^{5}+\frac{1}{x^5}$$
Given : $$x+\frac{1}{x}=3\sqrt2=k$$
Now, $$x^5+\frac{1}{x^5}=[(x^3+\frac{1}{x^3})\times(x^2+\frac{1}{x^2})]-(x+\frac{1}{x})$$
= $$[(x+\frac{1}{x})^3-3(x+\frac{1}{x})\times(x+\frac{1}{x})^2-2(x)(\frac{1}{x})]-(x+\frac{1}{x})$$
= $$[(k^3-3k)\times(k^2-2)]-(k)$$
= $$[(54\sqrt2-9\sqrt2)\times(18-2)]-(3\sqrt2)$$
= $$(45\sqrt2\times16)-3\sqrt2$$
= $$720\sqrt2-3\sqrt2=717\sqrt2$$
=> Ans - (D)
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