Question

if (a+b+c)(ab+bc+ca)=abc then prove that (a+b+c)^3=a^3+b^3+c^3​

Answer 1

Answer:

First we have:

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

Split (a+b+c) out of (a+b+c)^3:

(a+b+c)(a+b+c)^2 = (a^3+b^3+c^3)

Expand (a+b+c)^2:

(a+b+c)(a^2 + 2ab + b^2+ 2bc + c^2 + 2ca) = (a^3+b^3+c^3)

Rearrange into two groups:

2(a+b+c)(ab + bc + ca) + (a+b+c)(a^2+b^2+c^2) = (a^3+b^3+c^3)

Split (a+b+c)(a^2+b^2+c^2) into components:

2(a+b+c)(ab + bc + ca) + (a+ b+c)(a^2) +(a+b+c)(b^2) + (a+b +c)(c^2) = (a^3+b^3+c^3)

Obtain the cubes:

2(a+b+c)(ab + bc + ca) + (b+c)(a^2) +a^3 +(a+c)(b^2)+b^3 + (a+b)(c^2)+c^3 = (a^3+b^3+c^3)

Rearrange the cubes out:

2(a+b+c)(ab + bc + ca) + (b+c)(a^2) +(a+c)(b^2) + (a+b)(c^2) + (a^3+b^3+c^3) = (a^3+b^3+c^3)

Subtract out the cubes:

2(a+b+c)(ab + bc + ca) + (ab+ca)(a) +(ab+bc)(b) + (bc + ca)(c) = 0

Add 3 abc’s to both sides.

2(a+b+c)(ab + bc + ca) + (ab+ca)(a) + abc +(ab+bc )(b) +abc + (bc + ca)(c) + abc = 0 + abc+ abc+ abc

Complete (ab+bc+ca) for the remainder of the terms:

2(a+b+c)(ab + bc + ca) + (ab+ca + bc)(a) +(ab+bc + ca )(b) + (ab + bc + ca)(c) + abc = 3abc

Combine the terms:

3(a+b+c)(ab+bc+ca) = 3abc

Divide both sides by 3:

(a+b+c)(ab+bc+ca) = abc