Question

Prove that $(x^2+y^2{)}^4=(x^4-6x^2{y}^2+y^4{)}^2+(4x^{3}y-4x{y}^3{)}^2$.

Answer 1

$(x^2+y^2{)}^4={|x+iy|}^8$
$={\left|\right(x+iy{)}^2|}^4$
$={\left|\right(x^2-y^2)+2ixy)|}^4$
$={\left|\right[(x^2-y^2)+2ixy\left){]}^2\right|}^2$
$={\left|\right(x^2-y^2{)}^2+(2ixy{)}^2+2(x^2-y^2\left)\right(2ixy\left)\right|}^2$
$={|x^4+y^4-2x^2{y}^2-4x^2{y}^2+i(4x^{3}y-4x{y}^3\left)\right|}^2$
$={|x^4+y^4-6x^2{y}^2+i(4x^{3}y-4x{y}^3\left)\right|}^2$
$=(x^4-6x^2{y}^2+y^4{)}^2+(4x^{3}y-4x{y}^3{)}^2$