\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)-b^2\left(b-c\right)-b^2\left(a-b\right)+c^2\left(a-b\right)\)
\(=\left(a^2-b^2\right)\left(b-c\right)-\left(b^2-c^2\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(b-c\right)-\left(b-c\right)\left(b+c\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a+b-b-c\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
- Ta có: a2 (b-c)+b2 (c-a)+c2 (a-b)
= a2b - a2c + b2c - b2a + c2a - c2b
=a2b + b2c - a2c - c2b - b2a + c2a
=b(a2 + bc) - c(a2 + bc) - a(b2 - c2 )
=(a2 + bc)(b - c) - a(b - c)(b + c)
=(b - c)[a2 + bc - a(b + c)]
=(b-c)(a2 -ab -ac +bc)= (b-c)[a(a-b) - c(a-b)]
= (b - c)(a - b)(a - c)
- Vậy ...
