Question

(a+b+c) whole cube -a cube-b cube- c cube = 3 (a+b)(b+c)(c+a)...prove it..plzzzz i need it today only....very urgent

Answer 1

(a+b+c)³-a³-b³-c³
=(a+b)³+c³+3(a+b)c(a+b+c)-a³-b³-c³
[Using the formula, (a+b)³=a³+b³+3ab(a+b)]
=a³+b³+3ab(a+b)+3c(a+b)(a+b+c)-a³-b³
=3ab(a+b)+3c(a+b)(a+b+c)
=(a+b){3ab+3c(a+b+c)}
=3(a+b)(ab+ca+cb+c²)
=3(a+b){a(b+c)+c(b+c)}
=3(a+b)(b+c)(c+a) (Proved)

Answer 2

hope it will help you!
thanks