Ta có: \(\left(a+b+c\right)^3-\left(a+b-c\right)^3-\left(b+c-a\right)^3-\left(c+a-b\right)^3\)
\(=\left(a+b+c-a-b+c\right)\left[\left(a+b+c\right)^2+\left(a+b+c\right)\left(a+b-c\right)+\left(a+b-c\right)^2\right]-\left\{\left(b+c-a+c+a-b\right)\left[\left(b+c-a\right)^2+\left(b+c-a\right)\left(b-a-c\right)+\left(c+a-b\right)^2\right]\right\}\)
\(=2c\cdot\left[3a^2+3b^2+c^2+6ab\right]-2c\left(3a^2+3b^2+c^2-6ab\right)\)
\(=2c\left(3a^2+3b^2+c^2+6ab-3b^2-3a^2-c^2+6ab\right)\)
\(=2c\cdot12ab=24abc\)