Prove that

a³+b³ = (a+b) ³-3ab (a+b)

a³-b³ = (a-b) ³+3ab (a-b)

(For the first proof)

Start with (a+b) ^3

(a+b) ^3 = ((a+b) ^2) * (a+b)

or, (a+b) ^3 = (a^2+b^2+2ab) * (a+b)

or, (a+b) ^3=a^3+b*a^2+b^2*a+b^3+2a^2*b+2ab^2

or, (a+b) ^3=a^3+b^3+3a^2*b+3a*b^2

or, (a+b) ^3=a^3+b^3+3ab (a+b)

or, a^3+b^3 = (a+b) ^3-3ab (a+b)

(For the second proof)

Start with (a-b) ^3

(a-b) ^3 = ((a-b) ^2) * (a-b)

or, (a-b) ^3 = (a^2+b^2-2ab) * (a-b)

or, (a-b) ^3

=a^3-b*a^2+b^2*a-b^3-2a^2*b+2ab^2

or, (a-b) ^3=a^3-b^3-3a^2*b+3a*b^2

or, (a-b) ^3=a^3-b^3-3ab (a-b)

or, a^3-b^3 = (a-b) ^3+3ab (a-b)