If a, b and c be the roots of $x^3+p{x}^2+qx+r=0$, find the value of $(b+c-a)(c+a-b)(a+b-c)$

Given equation is $x^3+p{x}^2+qx+r=0$ Here, $a+b+c=-p,ab+bc+ca=q,abc=-r$

Now, 

$\begin{array}{l}(b+c-a)(c+a-b)(a+b-c)\\ =(b+c-a)(c+a-b)(a+b-c)\\ =(a+b+c-2a)(a+b+c-2b)(a+b+c-2c)\\ =(p-2a)(p-2b)(p-2c)\\ =p^3-2(a+b+c)p+2(ab+bc+ca)q-8abc\\ =p^3-2(-p)p+2\left(q\right)q-8(-r)\\ =p^3+2p^2+2q^2+8r\end{array}$