Đặt \(a+b=x,b+c=y,c+a=z\) ( đặt cho dễ nhìn ý mà :)) ta có :
\(\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(=x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-xz-yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(=2\left(a+b+c\right)\left[\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2-\left(a+b\right)\left(b+c\right)-\left(b+c\right)\left(c+a\right)-\left(c+a\right)\left(a+b\right)\right]\)
\(=2\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
Đặt \(\left(a+b;b+c;c+a\right)=\left(x;y;z\right)\)
\(x^3+y^3+z^3-3xyz=x^3+y^3+3xy\left(x+y\right)+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-yz-xz\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(=\frac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\right]\)
\(=\frac{1}{2}\left(2a+2b+2c\right)\left[\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\right]\)
\(=\left(a+b+c\right)\left[\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\right]\)