Question

Prove that
(a³+b³+c³)-a³-b³-c³=3(a+b)(b+c)(c+a).

Answer 1

Answer:

Step-by-step explanation:

(a+b+c)³-a³-b³-c³

=(a+b)³+3(a+b)²c+3(a+b)c²+c³-a³-b³-c³

=a³+3a²b+3ab²+b³+3(a²+2ab+b²)c+3ac²+3bc²+c³-a³-b³-c³

=3a²b+3ab²+3a²c+6abc+3b²c+3ac²+3bc²

=3(2abc+a²b+ab²+a²c+ac²+b²c+bc²)

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

=3(ab+b²+ac+bc)(c+a)

=3(abc+b²c+ac²+bc²+a²b+ab²+a²c+abc)

=3(2abc+a²b+ab²+a²c+ac²+b²c+bc²)

∴, LHS=RHS (Proved)

Answer 2

Step-by-step explanation:

a+b+c)³ = a³ + b³ + c³ + 3(a+b)(b+c)(c+a)

LHS = (a+b+c)³ - a³ - b³ - c³

= [a³ + b³ + c³ + 3(a+b)(b+c)(c+a)] - a³ - b³ - c³

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

= RHS

OR

(a+b+c)³-a³-b³-c³

=(a+b)³+3(a+b)²c+3(a+b)c²+c³-a³-b³-c³

=a³+3a²b+3ab²+b³+3(a²+2ab+b²)c+3ac²+3bc²+c³-a³-b³-c³

=3a²b+3ab²+3a²c+6abc+3b²c+3ac²+3bc²

=3(2abc+a²b+ab²+a²c+ac²+b²c+bc²)

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

=3(ab+b²+ac+bc)(c+a)

=3(abc+b²c+ac²+bc²+a²b+ab²+a²c+abc)

=3(2abc+a²b+ab²+a²c+ac²+b²c+bc²)

∴, LHS=RHS (Proved)