Question

(x+y)³+(x-y)³+x³-y³=?

Answer 1

(x+y)³ = x³ + y³ + 3x²y + 3xy²
(x-y)³ = x³ - y³ - 3x²y + 3xy²
thus

(x+y)³ + (x-y)³ + x³ - y³
=(x³ + y³ + 3x²y + 3xy²) + (x³ - y³ - 3x²y + 3xy²) + x³ - y³
=x³ + y³ + 3x²y + 3xy² +x³ - y³ - 3x²y + 3xy² + x³ - y³
= 3x³ - y³ + 6xy²

Answer 2

(x+y)^3+(x-y)^3+x^3-y^3
open the identity
(a+b)^3and (a-b)^3
(a+b)^3 = a^3+b^3+3a^2b+3ab^2
(a-b)^3 =a^3 +b^3-3a^2b+3ab^2
apply formula
{x^3+y^3+3x^2y+3xy^2} + { x^3+y^3-3x^2y+3xy^2} +x^3-y^3
3x^3+y^3+6xy^2
Answer