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\(=a^2b-a^2c+b^2c-b^2a+c^2a-c^2b\)
\(=\left(a^2b-b^2a\right)-\left(a^2c-b^2c\right)+c^2\left(a-b\right)\)
\(=ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)\)
\(=\left(a-b\right)\left(ab-ca-cb+c^2\right)\)
\(=\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
=a2b-a2c+b2c-ab2+ac2-bc2
=ab(a-b)-c(a-b)(a+b)+c2(a-b)
=(a-b)(ab-c+c2)
Bạn ơi bạn có thể ghi câu trả lời ra cụ thể giúp mình có được không ạ ?
\(a\left(b^2-c^2\right)-b\left(a^2-c^2\right)+c\left(a^2-b^2\right)\)
\(=ab^2-ac^2-ba^2+bc^2+ca^2-cb^2\)
\(=\left(ab^2-ac^2-bc^2\right)-\left(ba^2-bc^2-ca^2\right)\)
\(=a\left(b^2-c^2\right)-bc^2-a^2\left(b-c\right)+bc^2\)
\(=a\left(b^2-c^2\right)-a^2\left(b-c\right)\)
\(=a\left(b-c\right)\left(b+c\right)-a^2\left(b-c\right)\)
\(=\left(b+c\right)\left[a\left(b-c\right)-a^2\right]\)
\(=\left(b+c\right)\left(ab-ac-a^2\right)\)
\(a\left(b^2-c^2\right)-b\left(a^2-c^2\right)+c\left(a^2-b^2\right)\)
\(=c\left(a^2-b^2\right)+a\left(b^2-c^2\right)+b\left(c^2-a^2\right)\)
\(=-c\left[\left(b^2-c^2\right)+\left(c^2-a^2\right)\right]+a\left(b^2-c^2\right)+b\left(c^2-a^2\right)\)
\(=\left(a-c\right)\left(b^2-c^2\right)+\left(b-c\right)\left(c^2-a^2\right)\)
\(=\left(a-c\right)\left(b-c\right)\left(b+c\right)+\left(b-c\right)\left(c-a\right)\left(c+a\right)\)
\(=\left(a-c\right)\left(b-c\right)\left(b-a\right)\)
-c2(a - b) + b2(a - c) - a2(b - c)
= -c2a + c2b + b2a - b2c - a2b + a2c
= (a2c - c2a + c2b - b2c) + (b2a - a2b)
= c(a2 - ac + bc - b2) + ab(b - a)
= c2[(a - b)(a + b) - c(a - b)] - ab(a - b)
= c2(a - b)(a + b - c) - ab(a - b)
= (a - b)(c2a + c2b - c3 - ab)
a(b2 - c2) + b(c2-a2) + c(a2-b2)
=a(b2-c2) + b(a2-b2+b2-c2) + c(a2-b2)
=a(b2-c2) - b[(a2-b2)+(b2-c2)] + c(a2-b2)
=a(b2-c2) - b(a2-b2) - b(b2-c2) + c(a2-b2)
=[a(b2-c2) - b(b2-c2)] - [b(a2-b2)-c(a2-b2) ]
=(b2-c2)(a-b) - (a2-b2)(b-c)
=(a-b)(b-c)(b+c) - (a+b)(a-b)(b-c)
=(a-b)(b-c)[(b+c)-(a+b)]
=(a-b)(b-c)(c-a)