(b-a)*(c-a)*(c-b)*(c+b+a)

a(b+c)^2(b-c)+b(c+a)^2(C-a)+c(a+b)^2(a-b)

=a(b+c)^2(b-a)+bc(c+a)^2-ab(c+a)^2+ac(a+b^2-cb(a+b)^2

=a(b+c)^2(b-c)+cd(c-d)(2a+b+c)-a(bc^2+ba^2-ca^2-cb^2)

=(b-c)(ab^2+ac^2+2abc-2abc-cb^2-c^2b)a(b-c)(a^2-cb)

=(b-c)(ab^2+ac^2-ab^2-c^2b-a^3+abc)

=(b-c)(a-c)[b^2-(ac+a^2)+cb]

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

ap dung :(a-b-c)^2=a^2+b^2+c^2-2ab-2bc-2ca

ta dc:A=(a^2)^2+(b^2)^2+(c^2)^2-2.a^2.b^2-2.b^2-c^2-2.c^2.a^a

=>a=(a^2-b^2-c^2)^2

(a(b-c)^2 + b(c-a)^2 + c(a-b)^2) - (a^3 + b^3 + c^3) + 4abc

= a(b^2 - 2bc + c^2) + b(c^2 - 2ac + a^2) + c(a^2 - 2ab + b^2) - (a^3 + b^3 + c^3) + 4abc

= ab^2 - 2abc + ac^2 + bc^2 - 2abc + ba^2 + ca^2 - 2abc + cb^2 - a^3 - b^3 - c^3 + 4abc

= ab^2 + ac^2 + bc^2 + ba^2 + ca^2 + cb^2 - a^3 - b^3 - c^3 + 4abc - 6abc

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

= a^3 + b^3 + c^3 + a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 - a^3 - b^3 - c^3 - 2abc

= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 - 2abc

= ab(a + b) + ac(a + c) + bc(b + c) - 2abc

= (a + b)(ab - ac + bc) - 2abc

Vậy, ta có thể viết bài toán dưới dạng nhân tử là: (a + b)(ab - ac + bc) - 2abc.

\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)

\(=4a^2b^2-2ab\left(a^2+b^2-c^2\right)+2ab\left(a^2+b^2-c^2\right)-\left(a^2+b^2-c^2\right)^2\)

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

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

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

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

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

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

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

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

phân tích đa thức thành nhân tử

a^2(b-c)+b^2(c-a)+c^2(a-b)

= -(b-a)(c-a)(c-b)

nha bạn

a²(b-c)+b²(c-a)+c²(a-b)

=a²b-a²c+b²c-b²a+c²(a-b)

=(a²b-b²a)-(a²c-b²c)+c²(a-b)

=ab(a-b)+c(a²-b²)+c²(a-b)

=ab(a-b)+c(a-b)(a+b)+c²(a-b)

=(a-b)(ab+ac+bc+c²)

=(a-b)[(ab+bc)+(ac+c²)]

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

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

câu này mới đúng, câu trên mình sai

a²(b-c)+b²(c-a)+c²(a-b)

=a²b-a²c+b²c-b²a+c²(a-b)

=(a²b-b²a)-(a²c-b²c)+c²(a-b)

=ab(a-b)-c(a²-b²)+c²(a-b)

=ab(a-b)-c(a-b)(a+b)+c²(a-b)

=(a-b)(ab-ac-bc+c²)

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

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

Bạn thiếu dấu cộng à?

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

\(=ab^2-ac^2-ba^2+bc^2+ca^2-cb^2\)

\(=\left(ab^2-ac^2-bc^2\right)-\left(ba^2-bc^2-ca^2\right)\)

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

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

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

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

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

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

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

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

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

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

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