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# hard algebra!

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If x - 1/x = 2, then find the value of x^5 - 1/x^5.

Sep 8, 2020

By the binomial theorem, $$\left(x-\frac{1}{x}\right)^5=x^5-\frac{1}{x^5}-5x^3+\frac{5}{x^3}+10x-\frac{10}{x}=x^5-\frac{1}{x^5}-5(x^3-\frac{1}{x^3})+10(x-1/x)$$. Plugging, our expression is equivalent to $$x^5-\frac{1}{x^5}-5(x^3-\frac{1}{x^3})+20=32\rightarrow x^5-\frac{1}{x^5}-5(x^3-\frac{1}{x^3})=12$$. Now we just need to find x^3+1/x^3. By the binomial theorem, $$x^3-\frac{1}{x^3}-3(x+1/x)=8\rightarrow x^3-\frac{1}{x^3}=14$$. Plugging, $$x^5-\frac{1}{x^5}-70=12\rightarrow \boxed{x^5-\frac{1}{x^5}=82}$$