factor completely y^4-81x^4
\(y^4-81x^4=y^4-(3x)^4=(y^2-(3x)^2)(y^2+(3x)^2)=(y-3x)(y+3x)(y^2+9x^2)\)Depending on the field over which you're factoring, you may also write the last factor as\(y^2+9x^2=(y-3ix)(y+3ix)\)
