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1 tháng 6 2018

a) a2(a-b)-b2(a-c)-c2(b-a)

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

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

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

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

b)a(b-c)3+b(c-a)3+c(a-b)3

=a(b-c)3-b[(a-b)+(b-c)]+c(a-b)3

=a(b-c)3-b[(a-b)3+3(a-b)2(b-c)+3(a-b)(b-c)2+(b-c)3]+c(a-b)3

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

=(b-c)3(a-b)-(a-b)3(b-c)-3b(a-b)(b-c)(a-b+b-c)

=(b-c)3(a-b)-(a-b)3(b-c)-3b(a-b)(b-c)(a-c)

=(a-b)(b-c)[(b-c)2-(a-b)2-3b(a-c)]

=(a-b)(b-c)[(b-c-a+b)(b-c+a-b)-3b(a-c)]

=(a-b)(b-c)[(2b-a-c)(a-c)-3b(a-c)]

=(a-b)(b-c)(a-c)(2b-a-c-3b)

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

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

c)abc-(ab+ac+bc)+(a+b+c)-1

=abc-ab-ac-bc+a+b+c-1

=abc-bc-ab+b-ac+c+a-1

=bc(a-1)-b(a-1)-c(a-1)+a-1

=(a-1)(bc-b-c+1)

=(a-1)[b(c-1)-(c-1)]

=(a-1)(c-1)(b-1)

=(a-1)(b-1)(c-1)

18 tháng 9 2018

\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

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

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

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

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

18 tháng 9 2018

\(\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

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

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

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

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

 Châu ơi!đăng làm j z

31 tháng 7 2019

#)Giải :

a)\(ab\left(b-a\right)+bc\left(b-c\right)+ca\left(c-a\right)\)

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

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

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

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

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

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

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

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

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

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