Question

In any $\Delta ABC,$ prove that

$a^{3}sin(B-C)+b^{3}sin(C-A)+c^{3}sin(A-B)=0.$

Answer 1

By the sine rule, we have

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=k\left(say\right)$

$\Rightarrow a=ksinA,b=ksinBandc=ksinC.$

$a^{3}sin(B-C)=k^{3}si{n}^{3}A.sin(B-C)$

= $k^{3}si{n}^{2}A.sinA.sin(B-C)$

= $k^{3}si{n}^{2}A\left[sin\right[\pi -(B+C)\left]sin\right(B-C\left)\right][\therefore A+(B+c)=\pi ]$

= $k^{3}si{n}^{2}A\left[sin\right(B+C\left)sin\right(B-C\left)\right]$

= $k^{3}si{n}^{2}A.(si{n}^{2}B-si{n}^{2}C).$

similarly, $b^{3}sin(C-A)=k^{3}si{n}^{2}B(si{n}^{2}C-si{n}^{2}B).$

And, $c^{3}sin(A-B)=k^{3}si{n}^{2}C(si{n}^{2}A-si{n}^{2}B)$

$\therefore a^{3}sin(B-C)+b^{3}sin(C-A)+c^{3}sin(A-B)$

= $k^{3}si{n}^{2}A(si{n}^{2}B-si{n}^{2}C)+k^{3}si{n}^{2}B(si{n}^{2}C-si{n}^{2}A)+k^{3}si{n}^{2}C(si{n}^{2}A-si{n}^{2}B)$

= 0.