Without actually calculating the cubes,find t | vedclass

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(B) Let $x = \frac{1}{2}$,$y = \frac{1}{3}$,and $z = -\frac{5}{6}$.First,check the sum of $x, y,$ and $z$:$x + y + z = \frac{1}{2} + \frac{1}{3} - \fr

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$-\frac{5}{12}$

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(B) Let $x = \frac{1}{2}$,$y = \frac{1}{3}$,and $z = -\frac{5}{6}$.First,check the sum of $x, y,$ and $z$:$x + y + z = \frac{1}{2} + \frac{1}{3} - \frac{5}{6} = \frac{3+2-5}{6} = \frac{0}{6} = 0$.We know the algebraic identity: If $x + y + z = 0$,then $x^3 + y^3 + z^3 = 3xyz$.Substituting the values:$x^3 + y^3 + z^3 = 3 	imes \left(\frac{1}{2}ight) 	imes \left(\frac{1}{3}ight) 	imes \left(-\frac{5}{6}ight)$.$= 3 	imes \left(\frac{1}{6}ight) 	imes \left(-\frac{5}{6}ight) = 1 	imes \left(-\frac{5}{6}ight) = -\frac{5}{6}$.Wait,re-calculating: $3 	imes \frac{1}{2} 	imes \frac{1}{3} 	imes (-\frac{5}{6}) = 1 	imes (-\frac{5}{6}) = -\frac{5}{6}$.Correction: The expression is $3 	imes \frac{1}{2} 	imes \frac{1}{3} 	imes (-\frac{5}{6}) = \frac{3}{6} 	imes (-\frac{5}{6}) = \frac{1}{2} 	imes (-\frac{5}{6}) = -\frac{5}{12}$.

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

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