${sin}^{3} x + {sin}^{3} (60^{0} - x) - {sin}^{3} (60^{0} + x) \in$

[- 1, 1]

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[- \frac{1}{4}, \frac{1}{4}]

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[- \frac{3}{4}, \frac{3}{4}]

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[\frac{- 1}{2} \frac{1}{2}]

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Solution

The correct option is D $[- \frac{3}{4}, \frac{3}{4}]$
If $f x) = {sin}^{3} x + {sin}^{3} (60^{\circ} - x) - {sin}^{3} (60^{\circ} + x)$
We have, ${sin}^{3} A = \frac{3 sin A - sin 3 A}{4}$
Substituting in the given equation, we get
$\frac{3 sin x - sin 3 A}{4} + \frac{3 sin (60 - x) - sin 3 (60 - x)}{4} + \frac{3 sin (60 + x) - sin 3 (60 + x)}{4}$
U\sin g formula $sin A - sin B = 2 c o s (\frac{A + B}{2}) sin (\frac{A - B}{2}$ , we get
$= - \frac{3}{4} sin 3 x$
Range would be
$[- \frac{3}{4}, \frac{3}{4}]$
Hence, option 'C' is correct.

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sin3x+sin3(600−x)−sin3(600+x)∈

${sin}^{3} x + {sin}^{3} (60^{0} - x) - {sin}^{3} (60^{0} + x) \in$