AoPS Wiki talk:Problem of the Day/September 7, 2011

Solution

Raising that thing to the seventh doesn't look appealing, nor does solving for $x$ . So we try getting to $x^7+\frac{1}{x^7}$ gradually.

Since $x+\frac{1}{x}=1$ , $\left(x^2+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2=1$ and thus $x^2+\frac{1}{x^2}=-1$ .

We can also compute: $1(-1)=\left(x+\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}\right)=x^3+\frac{1}{x^3}+x+\frac{1}{x}=x^3+\frac{1}{x^3}+1$ so $x^3+\frac{1}{x^3}=-2$ .

Squaring $x^2+\frac{1}{x^2}$ , we see $1=\left(x^2+\frac{1}{x^2}\right)^2=x^4+\frac{1}{x^4}+2$ and thus $x^4+\frac{1}{x^4}=-1$ . Thus, $-1(-2)=\left(x^4+\frac{1}{x^4}\right)\left(x^3+\frac{1}{x^3}\right)=x^7+\frac{1}{x^7}+x+\frac{1}{x}=x^7+\frac{1}{x^7}+1$ and therefore, $x^7+\frac{1}{x^7}=2-1=\boxed{1}$ .

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AoPS Wiki talk:Problem of the Day/September 7, 2011

Solution