$$ A = \frac{(4\cdot2^4 + 1)(4\cdot4^4 + 1)(4\cdot6^4 + 1)}{(4\cdot1^4 + 1)(4\cdot3^4 + 1)(4\cdot7^4 + 1)}$$
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 $(1+4x^4)=(1-2x+2x^2)(1+2x+2x^2)$.
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.