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)

true

explanation:

30005

yes ok