Question

X∧4+4Y∧4X∧3-3X∧2-2X+216X∧2-9Y∧2+32X+16X∧4+X∧2Y∧2+Y∧4

Answer 1

x^4+4y^4

=x^4+4y^4+4x²y²-4x²y²

=(x²+2y²)²-4x²y²

=(x²+2y²+2xy)(x²+2y²-2xy)

X³-3X²-2X+2

=(x³+1)-(3x²+2x-1)

=(x+1)(x²-x+1)-(x+1)(3x-1)

=(x+1)(x²-x+1-3x+1)

=(x+1)(x²-4x+2)

16X²-9Y²+32X+16

=(4x+4)²-9y²

=(4x+4-3y)(4x+4+3y)

X^4+X²Y²+Y^4

=x^4+y^4+2x²y²-x²y²

=(x²+y²)²-x²y²

=(x²+y²+xy)(x²+y²-xy)