Question

Prove that (a+b)³+(b+c)³+(c+a)³-3(a+b)(b+c)(c+a)=2(a³+b³+c³-3abc).

Answer 1

Step-by-step explanation:  We are given to prove the following:

(a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)=2(a^3+b^3+c^3-3abc).

We will be using the following cubic formula in the proof:

(a+b)^3=a^3+b^3+3a^2b+3ab^2.

We have

L.H.S.\\\\=(a+b)^3+(b+c)^3+(c+a)^3-3(a+b)(b+c)(c+a)\\\\=a^3+b^3+3a^2b+3ab^2+b^3+c^3+3b^2c+3bc^2+c^3+a^3+3c^2a+3ca^2-3(ab+ac+b^2+bc)(c+a)\\\\=2a^3+2b^3+2c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3ca^2-3abc-3ac^2-3b^2c-3bc^2-3a^2b-3a^2c-3ab^2-3abc\\\\=2a^3+2b^3+2c^3-6abc\\\\=2(a^3+b^3+c^3-3abc)\\\\=R.H.S.

Hence proved.

Answer 2

Answer:

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

LHS = a³ + b³ + 3ab (a+b) + b³ + c³ + 3bc (b+c) + c³ + a³ + 3ca (c+a) - 3 (ab+ac+b²+bc) (c+a)

LHS = a³ + b³ + 3a²b + 3ab² + b³ + c³ + 3b²c + 3bc²+ c³+ a³ + 3c²a + 3ca²- 3(abc+a²b+ac²+a²c+b²c+b²a+bc²+abc)

LHS = 2a³ + 2b³ + 2c³ + 3a²b + 3ab² + 3b²c + 3bc² +3ac² + 3a²c - 6abc - 3a²b - 3ac² - 3a²c - 3b²c - 3b²a - 3bc².

LHS = 2 ( a³ + b³ + c³ - 3abc ) = RHS

HENCE PROVED