Question

Formula 1 bracket a + b + c cube equal a cube plus b cube plus c Cube + 3 bracket a + b bracket bracket b + c bracket C plus a proof

Answer 1

(a+b+c)^3 = {a+(b+c)}^3 = a^3 +(b+c)^3 + 3a(b+c){a+(b+c)} 
= a^3 +(b+c)^3 +3(ab+ac)(a+b+c) 
= a^3 +[b^3+c^3+3bc(b+c)]+3(ab+ac)(a+b+c) 
=(a^3+b^3+c^3) + 3bc(b+c) +3(ab+ac)(a+b+c) 
i.e. (a+b+c)^3 -(a^3+b^3+c^3)= 3bc(b+c)+3(ab+ac)(a+b+c) 
= 3b^2c+3bc^2 +3(a^2b+ab^2+abc+a^2c+abc+ac^2) 
=3[b^2c+bc^2+a^2b+ab^2+a^2c+ac^2+abc+abc... 
=3[(abc+ac^2+b^2c+bc^2) + (a^2b+a^2c+ab^2+abc)] 
=3[c(ab+ac+b^2+bc)+ a(ab+ac+b^2+bc)] 
= 3[(ab+ac+b^2+bc)(c+a)] 
=3[{a(b+c)+b(b+c)}(c+a)] 
=3[{(a+b)(b+c)}(c+a)] 
=3(a+b)(b+c)(c+a) proved