2021/002) It is given that the ratio of angles $A,\,B$ and $C$ is $1:2:4$ in a $\triangle ABC$, prove that $(a^2-b^2)(b^2-c^2)(c^2-a^2)=(abc)^2$.

We see that angles $A,\,B$ and $C$ are $\frac{\pi}{7}$, $\frac{2\pi}{7}$ and $\frac{4\pi}{7}$
and using laws of sines we need to prove

$(\sin ^2 A -\sin ^2 B)(\sin ^2 B - \sin ^2 C)(\sin ^2 C - \sin ^2 A)= (\sin\,A \sin\,B \sin\, C)^2$.
\where $A= \frac{\pi}{7}$ $B= \frac{2\pi}{7}$ and $C= \frac{4\pi}{7}$


to avoid fraction let $x =\frac{\pi}{7}$ so $A= x$ $B= 2x$ and $C= 4x$


so $\sin\,x = \sin 6x$ and $\sin 3x = \sin 4x$
we shall be using the following


$\sin\, X + \sin \,Y = 2 \sin \frac{X+Y}{2} \cos \frac{X-Y}{2}\cdots(1)$
$\sin\, X - \sin \,Y = 2 \sin \frac{X-Y}{2} \cos \frac{X+Y}{2}\cdots(2)$


So


$\sin ^2 X - \sin ^2 Y$
= $(\sin\, X + \sin \,Y)(\sin\, X - \sin \,Y)$
= $(2 \sin \frac{X+Y}{2} \cos \frac{X+Y}{2})(2 \sin \frac{X-Y}{2} \cos \frac{X-Y}{2})$
= $\sin (X+Y)\sin(X-Y)$


Now
LHS
= $(\sin ^2 A -\sin ^2 B)(\sin ^2 B - \sin ^2 C)(\sin ^2 C - \sin ^2 A)$
= $(\sin ^2 x -\sin ^2 2x)(\sin ^2 2x - \sin ^2 4x)(\sin ^2 4x - \sin ^2 x)$
= $(\sin ^2 2x -\sin ^2 x)(\sin ^2 4x - \sin ^2 2x)(\sin ^2 4x - \sin ^2 x)$ multiplying 1st and 2nd terms by -1
= $\sin 3x \sin \,x \sin 6x \sin 2x \sin 3x \sin \,x$
= $\sin 4x \sin \,x \sin x \sin 2x \sin 3x \sin \,x$ as $\sin 3x = \sin 4x$ and $\sin 6x =\sin \,x$
= $(\sin\,x \sin 2x \sin 4x)^2$
= $(\sin\,A \sin\,B \sin\, C)^2$
= RHS