Question

Without actually calculating the cubes, find the value of :

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

• A. $\frac{5}{12}$
• B. $-\frac{5}{12}$
• C. $\frac{7}{12}$
• D. $-\frac{7}{12}$

Answer 1

The correct option is B $-\frac{5}{36}$

We know that,

$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$

If $a+b+c=0$, then $a^3+b^3+c^3=3abc$

Now $\frac{1}{2}$ + $\frac{1}{3}$ - $\frac{5}{6}$

LCM of 2, 3, 6 is 6

Then $\frac{1}{2}$ + $\frac{1}{3}$ - $\frac{5}{6}$ = $\frac{3+2-5}{6}$

$\frac{1}{2}$ + $\frac{1}{3}$ - $\frac{5}{6}$ = 0

${\left(\frac{1}{2}\right)}^3$ + ${\left(\frac{1}{3}\right)}^3$ - ${\left(\frac{1}{3}\right)}^3$ = ($\frac{1}{2}$) ($\frac{1}{3}$)($\frac{-5}{6}$) = $\frac{-5}{36}$