Question

Factorise (a+b+c)³-a³-b³-c³​

Answer 1

Answer:

=[(a+b)+c]^3-a^3-b^3-c^3

=(a+b)^3+c^3-a^3-b^3-c^3

=(a+b)^3+c^3+3c(a+b)(a+b+c)-a^3-b^3-c^3

=a^3+b^3+3ab(a+b)+c^3+3c(a+b)(a+b+c)-a^3-b^3-c^3

=3ab(a+b)+3c(a+b)(a+b+c)

=3(a+b)[ab+c(a+b+c)]

=3(a+b)[ab+ac+bc+c^2]

=3(a+b)[a(b+c)+c(b+c)]

=3(a+b)(b+c)(a+c)