# Interesting 4th order factoring question

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

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?

You can use the standard factoring $(1+4x^4)=(1-2x+2x^2)(1+2x+2x^2)$.
Cancellations leave you with numerator 41,61 and denominator 113. Multiplying by $\frac{113}{61}$ leaves you with 41.
• You're welcome! Note that the factorisation is just a special case of $a^4+4b^4=(a^2-2ab+2b^2)(a^2+2ab+b^2)$. – almagest May 16 '16 at 19:00