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Algebra Difficulty 5.1 AIME, harder Find the answer

Compute 23−123+1⋅33−133+1⋅43−143+1⋅53−153+1⋅63−163+1 \frac{2^{3}-1}{2^{3}+1} \cdot \frac{3^{3}-1}{3^{3}+1} \cdot \frac{4^{3}-1}{4^{3}+1} \cdot \frac{5^{3}-1}{5^{3}+1} \cdot \frac{6^{3}-1}{6^{3}+1} .

A number or a short expression. Spacing and $ signs are ignored.

Solution

Use the factorizations n3−1=(n−1)(n2+n+1) n^{3}-1=(n-1)\left(n^{2}+n+1\right) and n3+1=(n+1)(n2−n+1) n^{3}+1=(n+1)\left(n^{2}-n+1\right) to write 1⋅73⋅3⋅2⋅134⋅7⋅3⋅215⋅13⋅4⋅316⋅21⋅5⋅437⋅31=1⋅2⋅433⋅6⋅7=4363 \frac{1 \cdot 7}{3 \cdot 3} \cdot \frac{2 \cdot 13}{4 \cdot 7} \cdot \frac{3 \cdot 21}{5 \cdot 13} \cdot \frac{4 \cdot 31}{6 \cdot 21} \cdot \frac{5 \cdot 43}{7 \cdot 31}=\frac{1 \cdot 2 \cdot 43}{3 \cdot 6 \cdot 7}=\frac{43}{63} .

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.