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Algebra Difficulty 5.2 AIME, harder Find the answer

10. (20 points) Let
f(x)=x5−3x3+2x2+3x+6,  f(x)=x^{5}-3 x^{3}+2 x^{2}+3 x+6 \text {, }

and let An=∏k=1n(4k−3)f(4k−3)(4k−1)f(4k−1)A_{n}=\prod_{k=1}^{n} \frac{(4 k-3) f(4 k-3)}{(4 k-1) f(4 k-1)}. Find the value of A25A_{25}.

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

Solution

10. Notice,
xf(x)=x(x5−3x3+2x2+3x+6)=x(x+2)(x2+x+1)(x2−3x+3)=(x3+3x2+3x+2)(x3−3x2+3x)=((x+1)3+1)((x−1)3+1)⇒An=∏k=1n((4k−2)3+1)((4k−4)3+1)((4k)3+1)((4k−2)3+1)=03+1(4n)3+1=164n3+1⇒A25=11000001. \begin{array}{l} x f(x)=x\left(x^{5}-3 x^{3}+2 x^{2}+3 x+6\right) \\ =x(x+2)\left(x^{2}+x+1\right)\left(x^{2}-3 x+3\right) \\ =\left(x^{3}+3 x^{2}+3 x+2\right)\left(x^{3}-3 x^{2}+3 x\right) \\ =\left((x+1)^{3}+1\right)\left((x-1)^{3}+1\right) \\ \Rightarrow A_{n}=\prod_{k=1}^{n} \frac{\left((4 k-2)^{3}+1\right)\left((4 k-4)^{3}+1\right)}{\left((4 k)^{3}+1\right)\left((4 k-2)^{3}+1\right)} \\ \quad=\frac{0^{3}+1}{(4 n)^{3}+1}=\frac{1}{64 n^{3}+1} \\ \Rightarrow A_{25}=\frac{1}{1000001} . \end{array}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.