Question

(a+b)(a^2-b^2)+(b+c)(b^2-c^2)+(c+a)(c^2-a^2)

Answer 1

(a+b)(a²-b²)+(b+c)(b²-c²)+(c+a)(c²-a²)

=(a+b)(a²-b²)+(b+c)[-(a²-b²)-(c²-a²)]+(c+a)(c²-a²)

=(a+b)(a²-b²)-(b+c)(a²-b²)-(b+c)(c²-a²)+(c+a)(c²-a²)

=(a²-b²)(a-c)-(a²-c²)(a-b)

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

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

=(a-b)(a-c)(b-c)

Answer 2

\(\left(a+b\right)\left(a^2-b^2\right)+\left(b+c\right)\left(b^2-c^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)

\(=a^3-ab^2+a^2b-b^3+b^3-bc^2+b^2c-c^3+c^3-a^2c+ac^2-a^3\)

\(=bc\left(b-c\right)+a^2\left(b-c\right)-a\left(b-c\right)\left(b+c\right)\)

\(=\left(b-c\right)\left(bc+a^2-ab-ac\right)\)

\(=\left(b-c\right)\left(a-c\right)\left(a-b\right)\)