Formula use:

\left(x-y\right)^3=x^3-y^3-3.x.y.\left(x-y\right)
\left(x+y\right)^3=x^3+3x^2y+3xy^2+y^3

Explanation

Here, given expression is :
\left(a+b+c\right)^3-\left(a-b-c\right)^3-6.\left(a+b+c\right)\left(a-b-c\right)\left(b+c\right)
Now, Simplify the expression :
\left(a+b+c\right)^3-\left(a-b-c\right)^3-6.\left(a+b+c\right)\left(a-b-c\right)\left(b+c\right)
\rightarrow\left(a+b+c\right)^3-\left(a-b-c\right)^3-3.\left(a+b+c\right)\left(a-b-c\right).2\left(b+c\right)
Let,
\left(a+b+c\right)=x
\left(a-b-c\right)=y
Now,
\left(x-y\right)
\rightarrow\{\left(a+b+c\right)-\left(a-b-c\right)\}
\rightarrow\left(a+b+c-a+b+c\right)
\rightarrow\left(2b+2c\right)
\therefore 2\left(b+c\right)
So that,
\left(a+b+c\right)^3-\left(a-b-c\right)^3-3.\left(a+b+c\right)\left(a-b-c\right).2\left(b+c\right)
\rightarrow\left(a+b+c\right)^3-\left(a-b-c\right)^3-3.\left(a+b+c\right)\left(a-b-c\right).\{\left(a+b+c\right)-\left(a-b-c\right)\}
\rightarrow x^3+y^3+3.x.y.\left(x-y\right)
\therefore\left(x-y\right)^3
Put the value of \left(x-y\right), we get :
\left(x-y\right)^3
\rightarrow\{2\left(b+c\right)\}^3
\rightarrow\left(2b+2c\right)^3
\rightarrow\left(2b\right)^3-3.\left(2b\right)^2.2c+3.2b.\left(2c\right)^2-\left(2c\right)^3
\rightarrow 8b^3-24b^2c+24bc^2-8c^3
\therefore 8\left(b^3-3b^2c+3bc^2-c^3\right)
Ans: 8\left(b^3-3b^2c+3bc^2-c^3\right)