Question

resolve into factors: p^4/q^4 + q^4/p^4 + 1

Answer 1

Answer:

$(\frac{p^2}{q^2} + \frac{q^2}{p^2} +1)(\frac{p^2}{q^2} + \frac{q^2}{p^2} -1)$ It was hard, so pls mark me as brainliest.

Step-by-step explanation:

$\frac{p^4}{q^4} + \frac{q^4}{p^4} + 1\\\frac{p^4}{q^4} + \frac{q^4}{p^4} + 2 - 1\\\\(\frac{p^4}{q^4} + \frac{q^4}{p^4} + 2) - 1\\((\frac{p^2}{q^2})^2 + (\frac{q^2}{p^2})^2 + 2(\frac{p^2}{q^2})(\frac{q^2}{p^2})) -1\\\\a^2 + b^2 + 2ab = (a + b)^2\\(\frac{p^2}{q^2} + \frac{q^2}{p^2} )^2 - 1\\(\frac{p^2}{q^2} + \frac{q^2}{p^2} )^2 - 1^2\\a^2 - b^2 = (a + b)(a - b)\\(\frac{p^2}{q^2} + \frac{q^2}{p^2} +1)(\frac{p^2}{q^2} + \frac{q^2}{p^2} -1)$