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## Homework Statement

[tex]sin ^{3}x + sin ^{3}(\frac{2\pi}{3} + x ) + sin ^{3}(\frac{4\pi}{3} + x) = \frac{-3sin3x}{4}[/tex]

## Homework Equations

[itex]sin 3x = 3sin 3x - 4sin^{3}x[/itex]

**Mod note**: The correct identity is ##sin(3x) = 3sin(x) - 4sin^3(x)##

The OP realized this earlier but was unable to edit his/her post.

## The Attempt at a Solution

I kind of felt that as there is cube and 3x, I should use the formula for [itex]sin 3x[/itex]. I multiplied and divided by 4. Then substituted from the identity. I got [itex]-3sinx[/itex] part but also got an additional term [itex]3sinx[/itex].

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