\(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Đặt a+b=x ; b+c=y; c+a=z ta có:
\(x^3+y^3+z^3-3xyz\)
=\(\left(x+y\right)^3-3x^2y-3xy^2+z^3-3xyz\)
=\(\left[\left(x+y\right)^3+z^3\right]-\left(3x^2y+3xy^2+3xyz\right)\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
xong thay vào