Without actually calculating the cubes, find | vedclass

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Question

Let $a=\frac{1}{2}, b=\frac{1}{3}, c=-\frac{5}{6}$

$	herefore \quad a+b+c=\frac{1}{2}+\frac{1}{3}-\frac{5}{6}$

$=\frac{3+2-5}{6}=\frac{0}{6}=0$

Vedclass · Accepted Answer

$-\frac{5}{12}$

Vedclass · Answer

Let $a=\frac{1}{2}, b=\frac{1}{3}, c=-\frac{5}{6}$

$	herefore \quad a+b+c=\frac{1}{2}+\frac{1}{3}-\frac{5}{6}$

$=\frac{3+2-5}{6}=\frac{0}{6}=0$

$\Rightarrow \quad a^{3}+b^{3}+c^{3}=3 a b c$

$	herefore \quad\left(\frac{1}{2}ight)^{3}+\left(\frac{1}{3}ight)^{3}-\left(\frac{5}{6}ight)^{3}=\left(\frac{1}{2}ight)^{3}+\left(\frac{1}{3}ight)^{3}+\left(-\frac{5}{6}ight)^{3}$

$=3 	imes \frac{1}{2} 	imes \frac{1}{3}\left(-\frac{5}{6}ight)=-\frac{5}{12}$

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 : $\left(\frac{1}{2}\right)^{3}+\left(\frac{1}{3}\right)^{3}-\left(\frac{5}{6}\right)^{3}$