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$$ A = \frac{(4\cdot2^4 + 1)(4\cdot4^4 + 1)(4\cdot6^4 + 1)}{(4\cdot1^4 + 1)(4\cdot3^4 + 1)(4\cdot7^4 + 1)}$$

What is the value of $ \dfrac{113A}{61}$ ?

So i tried factoring this $\dfrac{(4\cdot(x+1)^4 + 1)(4\cdot(x+3)^4 + 1)(4(x+5)^4 + 1)}{(4x^4 + 1)(4(x+2)^4 + 1)(4(x+6)^4 + 1)}$ in Wolframalpha

And it gave $\dfrac{(2x^2 + 14x + 25)(2x^2 + 18x + 41)}{(2x^2 + 26x + 85)(2x^2 -2x + 1)}$

And we can give x=1 and find answer but i need a easier way. Can you help me?

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You can use the standard factoring $(1+4x^4)=(1-2x+2x^2)(1+2x+2x^2)$.

Applying it to the numerator you get the factors: 5,13; 41,25; 61,85.

Applying it to the denominator you get the factors: 5; 13,25; 85,113.

Cancellations leave you with numerator 41,61 and denominator 113. Multiplying by $\frac{113}{61}$ leaves you with 41.

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  • $\begingroup$ You're welcome! Note that the factorisation is just a special case of $a^4+4b^4=(a^2-2ab+2b^2)(a^2+2ab+b^2)$. $\endgroup$ – almagest May 16 '16 at 19:00

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