tk mình đi mình giải cho 

\(B=\left(2xy-x^2-y^2+z^2\right)\left(2xy+x^2+y^2-z^2\right)\)

\(=\left[z^2-\left(x-y\right)^2\right]\left[\left(x+y\right)^2-z^2\right]\)

\(=\left(z-x+y\right)\left(z+x-y\right)\left(x+y-z\right)\left(x+y+z\right)\)

Câu hỏi của Access_123 - Toán lớp 8 - Học toán với OnlineMath

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

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

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

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

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

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

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

ab(a-b)+bc(b-c)+ca(c-a)

= ab(a-b)+bc(b-a+a-c)+ca(c-a)

=ab(a-b)+bc(b-a)+bc(a-c)+ca(c-a)

=ab(a-b)-bc(a-b)-bc(c-a)+ca(c-a)

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

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

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

a^3+b^3+c^3-3abc

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

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

=(a+b+c)(a^2+2ab+b^2-bc-ca+c^2)-3ab(a+b+c)

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