Competition problem — Algebra — Maths Olympiad Prep

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Algebra Difficulty 4.9 AIME Find the answer

Let $r=H_{1}$ be the answer to this problem. Given that $r$ is a nonzero real number, what is the value of $r^{4}+4 r^{3}+6 r^{2}+4 r ?$

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Since $H_{1}$ is the answer, we know $r^{4}+4 r^{3}+6 r^{2}+4 r=r \Rightarrow(r+1)^{4}=r+1$ . Either $r+1=0$ , or $(r+1)^{3}=1 \Rightarrow r=0$ . Since $r$ is nonzero, $r=-1$ .

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