Question

Factorise (x3-y3)^3+(y3-z3)^3+(z3-x3)^3

Answer 1

Answer:

81(x^2 z -x^2 y +x y^2 -x z^2 +y x^2 -y^2 z)

Step-by-step explanation:

=[(x)(3-y)(3)]^3 +[(y)(3-z)(3)]^3 +[(z)(3-x)(3)]^3

=27(x-y)^3 +27(y-z)^3 +27(z-x)^3

factor out the common term 27:

=27 [(x-y)^3+(y-z)^3+(-x+z)^3]

Factor (x-y)^3+(y-z)^3+(z-x)^3: 3(-x^2 y +x^2 z +x y^2 -x z^2 +y z^2 -y^2 z

=27(3)(x^2 z -x^2 y +x y^2 -x z^2 +y z^2 -y^2 z)

=81(x^2 z -x^2 y +x y^2 -x z^2 +y x^2 -y^2 z)