Answer :
so we kkonw that 64=4^3
so if we factor out x we get
x(4^3(x^3)+y^3) or x[(4x)^3+y^3)
so the additon of two perfect cubes so
(4x+y)(16x^2+4xy+y^2)
and we remember that this is all times x so
x[(4x+y(16^2+4xy+y^2))]
or (x)(4x+y)(16x^2+4xy+y^2)
so if we factor out x we get
x(4^3(x^3)+y^3) or x[(4x)^3+y^3)
so the additon of two perfect cubes so
(4x+y)(16x^2+4xy+y^2)
and we remember that this is all times x so
x[(4x+y(16^2+4xy+y^2))]
or (x)(4x+y)(16x^2+4xy+y^2)
