\(a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

\(=\left(a+b+c\right)^3\)

Cảm ơn Nguyễn Đức Anh :))

\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)

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

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

\(=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)\)

\(=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2-ab+b^2\right)\)

\(=\left(a-b\right)\left(-a^2b-ab^2-c^3+a^2c-abc+b^2c\right)\)

(a(b-c)^2 + b(c-a)^2 + c(a-b)^2) - (a^3 + b^3 + c^3) + 4abc

= a(b^2 - 2bc + c^2) + b(c^2 - 2ac + a^2) + c(a^2 - 2ab + b^2) - (a^3 + b^3 + c^3) + 4abc

= ab^2 - 2abc + ac^2 + bc^2 - 2abc + ba^2 + ca^2 - 2abc + cb^2 - a^3 - b^3 - c^3 + 4abc

= ab^2 + ac^2 + bc^2 + ba^2 + ca^2 + cb^2 - a^3 - b^3 - c^3 + 4abc - 6abc

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

= a^3 + b^3 + c^3 + a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 - a^3 - b^3 - c^3 - 2abc

= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 - 2abc

= ab(a + b) + ac(a + c) + bc(b + c) - 2abc

= (a + b)(ab - ac + bc) - 2abc

Vậy, ta có thể viết bài toán dưới dạng nhân tử là: (a + b)(ab - ac + bc) - 2abc.