hard algebra!

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, \(x^5-\frac{1}{x^5}-70=12\rightarrow \boxed{x^5-\frac{1}{x^5}=82}\)