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(C) Let $a = \frac{1}{2}$,$b = \frac{1}{3}$,and $c = -\frac{5}{6}$.First,calculate the sum $a + b + c$:$a + b + c = \frac{1}{2} + \frac{1}{3} - \frac{

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$-\frac{5}{12}$

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(C) Let $a = \frac{1}{2}$,$b = \frac{1}{3}$,and $c = -\frac{5}{6}$.First,calculate the sum $a + b + c$:$a + b + c = \frac{1}{2} + \frac{1}{3} - \frac{5}{6} = \frac{3 + 2 - 5}{6} = \frac{0}{6} = 0$.We know the algebraic identity: If $a + b + c = 0$,then $a^{3} + b^{3} + c^{3} = 3abc$.Applying this identity to the given expression:$\left(\frac{1}{2}ight)^{3} + \left(\frac{1}{3}ight)^{3} + \left(-\frac{5}{6}ight)^{3} = 3 	imes \left(\frac{1}{2}ight) 	imes \left(\frac{1}{3}ight) 	imes \left(-\frac{5}{6}ight)$.$= 3 	imes \frac{1}{6} 	imes \left(-\frac{5}{6}ight) = \frac{1}{2} 	imes \left(-\frac{5}{6}ight) = -\frac{5}{12}$.

Without actually calculating the cubes,find the value of:
$\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}$

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Without actually calculating the cubes,find the value of:$\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}$