In Δ A B C ,   a 3 cos B - C + b 3 cos C - A + c 3 cos A - B =

We know that, $\sin 2x+\sin 2y=2\sin \left(x+y\right)\cos \left(x-y\right)$

$\Rightarrow \cos \left(x-y\right)=\frac{\sin 2x+\sin 2y}{2\sin \left(x+y\right)}$

$\Rightarrow \cos \left(x-y\right)=\frac{2\sin x\cos x+2\sin y\cos y}{2\sin \left(x+y\right)}$

$\Rightarrow \cos \left(x-y\right)=\frac{\sin x\cos x+\sin y\cos y}{\sin \left(x+y\right)}$

Now, $a^3\cos \left(B-C\right)=a^3\left[\frac{\sin B\cos B+\sin C\cos C}{\sin \left(B+C\right)}\right]$

$\Rightarrow a^3\cos \left(B-C\right)=a^3\left[\frac{\sin B\cos B+\sin C\cos C}{\sin \left(\mathrm{\pi }-\mathrm{A}\right)}\right]$

$\Rightarrow a^3\cos \left(B-C\right)=a^3\left[\frac{\sin B\cos B+\sin C\cos C}{\mathrm{sinA}}\right]$

Now

$a^3\cos \left(B-C\right)=a^3\left[\frac{bk\cos B+ck\cos C}{ak}\right]$

                             $a^3\cos \left(B-C\right)=a^2\left(b\cos B+c\cos C\right) ...\left(\mathrm{i}\right)$

Similarly

$b^3\cos \left(C-A\right)=b^2\left(c\cos C+a\cos A\right) ...\left(\mathrm{ii}\right)$

$c^3\cos \left(A-B\right)=c^2\left(a\cos A+b\cos B\right) ...\left(\mathrm{iii}\right)$

Adding $\left(\mathrm{i}\right), \left(\mathrm{ii}\right) \& \left(\mathrm{iii}\right)$, we get

$a^3\cos (B-C)+b^3\cos (C-A)+c^3\cos (A-B)$

$=a^2\left(b\cos B+c\cos C\right)+b^2\left(a\cos A+c\cos C\right)+c^2\left(b\cos B+a\cos A\right)$

$=ab\left(a\cos B+b\cos A\right)+ac\left(a\cos C+c\cos A\right)+bc\left(b\cos C+c\cos B\right)$

$=3abc; \left[\because a\cos B+b\cos A=c\right]$ using projection formula.