Prove that: \[ \sin^3 x+\sin^3\left(\frac{2\pi}{3}+x\right) +\sin^3\left(\frac{4\pi}{3}+x\right) = -\frac{3}{4}\sin 3x \]
Solution
Using the identity
\[
\sin^3\theta
=
\frac{3\sin\theta-\sin3\theta}{4}
\]
we get
\[
\sin^3 x
=
\frac{3\sin x-\sin3x}{4}
\]
\[
\sin^3\left(\frac{2\pi}{3}+x\right)
=
\frac{
3\sin\left(\frac{2\pi}{3}+x\right)
–
\sin\left(2\pi+3x\right)
}{4}
\]
\[
\sin^3\left(\frac{4\pi}{3}+x\right)
=
\frac{
3\sin\left(\frac{4\pi}{3}+x\right)
–
\sin\left(4\pi+3x\right)
}{4}
\]
Since
\[
\sin(2\pi+3x)=\sin3x
\]
\[
\sin(4\pi+3x)=\sin3x
\]
Adding all three expressions,
\[
LHS
=
\frac{
3\sin x
+3\sin\left(\frac{2\pi}{3}+x\right)
+3\sin\left(\frac{4\pi}{3}+x\right)
-3\sin3x
}{4}
\]
Now use the identity
\[
\sin x
+\sin\left(\frac{2\pi}{3}+x\right)
+\sin\left(\frac{4\pi}{3}+x\right)
=0
\]
Therefore,
\[
LHS
=
\frac{-3\sin3x}{4}
\]
\[
=
-\frac{3}{4}\sin3x
\]
Hence proved.