Question

In a $△ABC$, $a$, $b$, $c$ are the sides of the triangle opposite to the angles $A$, $B$, $C$ respectively. Then, the value of $a^3\sin \left(B-C\right)+b^3\sin \left(C-A\right)+c^3\sin \left(A-B\right)$ is

• A. $0$
• B. $1$
• C. $3$
• D. $2$

Answer 1

The correct option is A

$0$

Explanation for the correct option:

It is given that in a $△ABC$, $a$, $b$, $c$ are the sides of the triangle opposite to the angles $A$, $B$, $C$ respectively, as shown in the figure below.

It is known by the sine rule that $\frac{\mathbf{a}}{\mathbf{sinA}}\mathbf{=}\frac{\mathbf{b}}{\mathbf{sinB}}\mathbf{=}\frac{\mathbf{c}}{\mathbf{sinC}}$.

Let us assume that $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k..\left(1\right)$.

From the equation $\left(1\right)$, $a=k\sin A$.

$$\begin{gathered} \Rightarrow a^3=k^3{\sin }^{3}A \\ \Rightarrow a^3=k^3{\sin }^3\left[\mathrm{\pi }-\left(\mathrm{B}+\mathrm{C}\right)\right]\left[\because A+B+C=\mathrm{\pi }\right] \\ \Rightarrow a^3=k^3{\sin }^3\left(B+C\right) \end{gathered}$$

Thus, $a^3\sin \left(B-C\right)=k^3{\sin }^3\left(B+C\right)\sin \left(B-C\right)$

$$\begin{gathered} \Rightarrow a^3\sin \left(B-C\right)=k^3{\sin }^2\left(B+C\right)\sin \left(B+C\right)\sin \left(B-C\right) \\ \Rightarrow a^3\sin \left(B-C\right)=k^3{\sin }^2\left(\mathrm{\pi }-\mathrm{A}\right)\left({\sin }^{2}B-{\sin }^{2}C\right)\mathbf{}\left[\because {\mathbf{sin}}^{\mathbf{2}}\mathit{\alpha }\mathbf{-}{\mathbf{sin}}^{\mathbf{2}}\mathit{\beta }\mathbf{=}\mathbf{sin}\mathbf{(}\mathbf{\alpha }\mathbf{+}\mathbf{\beta }\mathbf{)}\mathbf{sin}\mathbf{(}\mathbf{\alpha }\mathbf{-}\mathbf{\beta }\mathbf{)}\right] \\ \Rightarrow a^3\sin \left(B-C\right)=k^3{\sin }^{2}A\left({\sin }^{2}B-{\sin }^{2}C\right)\mathbf{[}\mathbf{\because }\mathbf{s}\mathbf{i}\mathbf{n}\left(\pi -\alpha \right)\mathbf{=}\mathbf{s}\mathbf{i}\mathbf{n}\left(\alpha \right)\mathbf{]} \end{gathered}$$

From the equation $\left(1\right)$, $b=k\sin B$.

$$\begin{gathered} \Rightarrow b^3=k^3{\sin }^{3}B \\ \Rightarrow b^3=k^3{\sin }^3\left[\mathrm{\pi }-\left(A+\mathrm{C}\right)\right]\left[\because A+B+C=\mathrm{\pi }\right] \\ \Rightarrow b^3=k^3{\sin }^3\left(A+C\right) \end{gathered}$$

Thus, $b^3\sin \left(C-A\right)=k^3{\sin }^3\left(C+A\right)\sin \left(C-A\right)$

$$\begin{gathered} \Rightarrow b^3\sin \left(C-A\right)=k^3{\sin }^2\left(C+A\right)\sin \left(C+A\right)\sin \left(C-A\right) \\ \Rightarrow b^3\sin \left(C-A\right)=k^3{\sin }^2\left(\mathrm{\pi }-B\right)\left({\sin }^{2}C-{\sin }^{2}A\right)\left[\because {\mathbf{sin}}^{\mathbf{2}}\mathit{\alpha }\mathbf{-}{\mathbf{sin}}^{\mathbf{2}}\mathit{\beta }\mathbf{=}\mathbf{sin}\mathbf{(}\mathbf{\alpha }\mathbf{+}\mathbf{\beta }\mathbf{)}\mathbf{sin}\mathbf{(}\mathbf{\alpha }\mathbf{-}\mathbf{\beta }\mathbf{)}\right] \\ \Rightarrow b^3\sin \left(C-A\right)=k^3{\sin }^{2}B\left({\sin }^{2}C-{\sin }^{2}A\right)\mathbf{[}\mathbf{\because }\mathbf{sin}\mathbf{(}\mathbf{\pi }\mathbf{-}\mathbf{\alpha }\mathbf{)}\mathbf{=}\mathbf{sin}\mathbf{\left(}\mathbf{\alpha }\mathbf{\right)}\mathbf{]} \end{gathered}$$

From the equation $\left(1\right)$, $c=k\sin C$.

$$\begin{gathered} \Rightarrow c^3=k^3{\sin }^{3}C \\ \Rightarrow c^3=k^3{\sin }^3\left[\mathrm{\pi }-\left(\mathrm{A}+\mathrm{B}\right)\right]\left[\because A+B+C=\pi \right] \\ \Rightarrow c^3=k^3{\sin }^3\left(A+B\right) \end{gathered}$$

Thus, $c^3\sin \left(A-B\right)=k^3{\sin }^3\left(A+B\right)\sin \left(A-B\right)$

$$\begin{gathered} \Rightarrow c^3\sin \left(A-B\right)=k^3{\sin }^2\left(A+B\right)\sin \left(A+B\right)\sin \left(A-B\right) \\ \Rightarrow c^3\sin \left(A-B\right)=k^3{\sin }^2\left(\mathrm{\pi }-C\right)\left({\sin }^{2}A-{\sin }^{2}B\right)\left[\because {\mathbf{sin}}^{\mathbf{2}}\mathit{\alpha }\mathbf{-}{\mathbf{sin}}^{\mathbf{2}}\mathit{\beta }\mathbf{=}\mathbf{sin}\mathbf{(}\mathbf{\alpha }\mathbf{+}\mathbf{\beta }\mathbf{)}\mathbf{sin}\mathbf{(}\mathbf{\alpha }\mathbf{-}\mathbf{\beta }\mathbf{)}\right] \\ \Rightarrow c^3\sin \left(A-B\right)=k^3{\sin }^{2}C\left({\sin }^{2}A-{\sin }^{2}B\right)\mathbf{[}\mathbf{\because }\mathbf{sin}\mathbf{(}\mathbf{\pi }\mathbf{-}\mathbf{\alpha }\mathbf{)}\mathbf{=}\mathbf{sin}\mathbf{\left(}\mathbf{\alpha }\mathbf{\right)}\mathbf{]} \end{gathered}$$

Thus, $a^3\sin \left(B-C\right)+b^3\sin \left(C-A\right)+c^3\sin \left(A-B\right)=k^3{\sin }^{2}A\left({\sin }^{2}B-{\sin }^{2}C\right)+k^3{\sin }^{2}B\left({\sin }^{2}C-{\sin }^{2}A\right)+k^3{\sin }^{2}C\left({\sin }^{2}A-{\sin }^{2}B\right)$

$$\begin{gathered} \Rightarrow a^3\sin \left(B-C\right)+b^3\sin \left(C-A\right)+c^3\sin \left(A-B\right)=k^3\left({\sin }^{2}A·{\sin }^{2}B-{\sin }^{2}A·{\sin }^{2}C+{\sin }^{2}B·{\sin }^{2}C-{\sin }^{2}B·{\sin }^{2}A+{\sin }^{2}C·{\sin }^{2}A-{\sin }^{2}C·{\sin }^{2}B\right) \\ \Rightarrow a^3\sin \left(B-C\right)+b^3\sin \left(C-A\right)+c^3\sin \left(A-B\right)=k^3\times 0 \\ \Rightarrow a^3\sin \left(B-C\right)+b^3\sin \left(C-A\right)+c^3\sin \left(A-B\right)=0 \end{gathered}$$Therefore, $a^3\sin \left(B-C\right)+b^3\sin \left(C-A\right)+c^3\sin \left(A-B\right)=0$.

Hence, option(A) is the correct option.