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=a3(b-c)-b3(a-c)+c3(a-c-b+c)
=a3(b-c)-b3(a-c)+c3(a-c)-c3(b-c)
=(a3-c3)(b-c)-(b3-c3)(a-c)
=(a-c)(a2+ac+c2)(b-c)-(b-c)(b2+bc+c2)(a-c)
=(b-c)(a-c)(a2+ac+c2-b2-bc-c2)
=(b-c)(a-c)[(a-b)(a+b)+c(a-b)]
=(a-b)(b-c)(a-c)(a+b+c)
Ta có:\(a^3\left(b-c\right)+b^3\left(c-a\right)+c^3\left(a-b\right)\)
\(=a^3\left(b-c\right)-b^3\left(a-c\right)+c^3\left(a-c-b+c\right)\)
\(=a^3\left(b-c\right)-b^3\left(a-c\right)+c^3\left(a-c\right)-c^3\left(b-c\right)\)
\(=\left(a^3-c^3\right)\left(b-c\right)-\left(b^3-c^3\right)\left(a-c\right)\)
\(=\left(a-c\right)\left(a^2+ac+c^2\right)\left(b-c\right)-\left(b-c\right)\left(b^2+bc+c^2\right)\left(a-c\right)\)
\(=\left(b-c\right)\left(a-c\right)\left(a^2+ac+c^2-b^2-bc-c^2\right)\)
\(=\left(b-c\right)\left(a-c\right)\left[\left(a-b\right)\left(a+b\right)+c\left(a-b\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)
\(a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)\)
\(=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c\)
\(=\left(ab^3-a^3b\right)+\left(bc^3-ac^3\right)+\left(a^3c-b^3c\right)\)
\(=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)\)
\(=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2-ab+b^2\right)\)
\(=\left(a-b\right)\left(-a^2b-ab^2-c^3+a^2c-abc+b^2c\right)\)
Đặt \(a-b=x\) , \(b-c=y\) và \(c-a=z\)
\(\Rightarrow x+y+z=\left(a-b\right)+\left(b-c\right)+\left(c-a\right)=0\)
Chắc bạn cùng biết \(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)
Vậy \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Chúc bạn học tốt.
a(b^3-c^3)+b(c^3-a^3)+c(a^3-b^3) =a(b^3-a^3+a^3-c^3)+b(c^3-a^3)+c(a^3-b^3) = -a(a^3-b^3)-a(c^3-a^3)+b(c^3-a^3)+c(a^3-b^3) = (a^3-b^3)(c-a)-(c^3-a^3)(a-b) = (a-b)(c-a)(a^2+ab+b^2)-(a-b)(c-a)(c^2+ac+b^2) = (a-b)(c-a)(a^2+ab+b^2-c^2-ac-b^2) = (a-b)(c-a)(a^2+ab-c^2-ac)
Có vẻ nhìn hơi rối mắt bạn thông cảm nha
\(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)
\(=a^3-3a^2b+3ab^2-b^3+b^3-3b^2c+3bc^2-c^3+c^3-3c^2a+3ca^2-a^3\)
\(=-3a^2b+3ab^2-3b^2c+3bc^2-3c^2a+3ca^2\)
= -3ab (a-b) - 3bc(b-c) - 3ca(c-a)