Question

Let $ABC$ be a triangle with sides $a,b,c$ and corresponding angles $A,B,C$ respectively. If angle $A=3B$, then $(a^2-b^2)(a-b)$ is equal to

• A. $b{c}^2$
• B. $a^{2}b$
• C. $a{b}^2$
• D. $b^{2}c$

Answer 1

The correct option is A $b{c}^2$

We know that $a=2R\sin A$
Now, $(a^2-b^2)(a-b)$
$=8R^3({\sin }^{2}A-{\sin }^{2}B)(\sin A-\sin B)$
$=8R^3({\sin }^{2}3B-{\sin }^{2}B)(\sin 3B-\sin B)(\because A=3B)$
$=8R^3(\sin 3B-\sin B{)}^2(\sin 3B+\sin B)$
$=8R^3\left(2\cos 2B\sin B{)}^2\right(2\sin 2B\cos B)$
$=8R^3\left[\sin 4B\right(2\sin 2B\cos 2B\left)\sin B\right]$
$=8R^3{\sin }^{2}4B\sin B$
$=8R^3{\sin }^2(\pi -C)\sin B(\because 3B+B+C=\pi )$
$=\left(2R\sin C{)}^2\right(2R\sin B)$
$\therefore (a^2-b^2)(a-b)=b{c}^2$