Answer
$3x^2$.
Work Step by Step
The given expression is
$\frac{x^4-y^8}{x^2+y^4}\div \frac{x^2-y^4}{3x^2}$
Replace divide sign with multiplication sign and reciprocal the right hand side fraction as shown below.
$\frac{x^4-y^8}{x^2+y^4}\times \frac{3x^2}{x^2-y^4}$
Factor each denominator and numerator.
$=x^4-y^8$
Use algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=(x^2)^2-(y^4)^2$
$=(x^2+y^4)(x^2-y^4)$
Again use algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=(x^2+y^4)(x^2-(y^2)^2)$
$=(x^2+y^4)(x+y^2)(x-y^2)$
$x^2+y^4$
This is the prime polynomial.
$=x^2-y^4$
Use algebraic identity $a^2-b^2=(a+b)(a-b)$.
$=x^2-(y^2)^2$
$=(x+y^2)(x-y^2)$
$=x+y$
This is the prime polynomial.
$3x^2$
This is the prime polynomial.
Substitute all factors into the given expression.
$\frac{(x^2+y^4)(x+y^2)(x-y^2)}{x^2+y^4}\times \frac{3x^2}{(x+y^2)(x-y^2)}$
Cancel common factors.
The remaining fraction is
$=3x^2$.
