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29 tháng 10 2017

a(b2+c2)+b(c2+a2)+c(a2+b2)+22abc

= ab2+ac2+bc2+a2b+(a2c+b2c+2abc)

= ab(a+b)+c2(a+b)+c(a+b)2

= (a+b)(ab+c2+ac+bc)

= (a+b)[a(b+c)+c(b+c)

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

b)

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

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

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

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

= (a-b)(a2-ac+ab-bc +c2-a2)

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

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

29 tháng 10 2017

\(a^3\left(c-b^2\right)+b^3\left(a-c^2\right)+c^3\left(b-a^2\right)+abc\left(abc-1\right)\\ =a^3\left(c-b^2\right)+ab^3-b^3c^2+bc^3-a^2c^3+a^2b^2c^2-abc\\ =a^3\left(c-b^2\right)+bc^2\left(c-b^2\right)-ab\left(c-b^2\right)-a^2c^2\left(c-b^2\right)\\ =\left(c-b^2\right)\left(a^3+bc^2-ab-a^2c^2\right)\\ =\left(c-b^2\right)\left[a^2\left(a-c^2\right)-b\left(a-c^2\right)\right]\\ =\left(c-b^2\right)\left(a-c^2\right)\left(a^2-b\right)\)

1 tháng 10 2016

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

21 tháng 9 2016

a3(c - b2) + b3(a - c2) + c3(b - a2) + abc(abc - 1)

= a3c - a3b2 + ab3 - b3c2 + bc3 - a2c3 + a2b2c2 - abc

= a2b2c2 - b3c2 - (a2c3 - bc3) - (a3b2 - ab3) + (a3c - abc)

= b2c2(a2 - b) - c3(a2 - b) - ab2(a2 - b) + ac(a2 - b)

= (a2 - b)(b2c2 - c3 - ab2 + ac) = (a2 - b)[c2(b2 - c) - a(b2 - c)] = (a2 - b)(b2 - c)(c2 - a)

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

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

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

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

b: \(=ab^2+ac^2+bc^2+a^2b+a^2c+b^2c+2abc\)

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

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

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

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

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

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

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

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

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

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

29 tháng 9 2018

a3(c - b2) + b(a - c2) + c3(b - a2) + abc(abc - 1)

= a3c - a3b2 + ab3 - b3c2 + c3b - a2c3 + a2b2c2 - abc

= (a2b2c2 - b3c2) + (a3c - abc) - (a3b2 - ab3) - (a2c3 - c3b)

= b2c2(a2 - b) + ac(a2 - b) - ab2(a2 - b) - c3(a2 - b)

= (a2 - b)(b2c2 + ac - ab2 - c3)

= (a2 - b)[(b2c2 - c3) - (ab2 - ac)]

= (a2 - b)[c2(b2 - c) - a(b2 - c)]

= (a2 - b)(c2 - a)(b2 - c)