Question

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

• A. $\frac{=5}{6}$
• B. $\frac{-5}{12}$
• C. $\frac{4}{5}$
• D. $\frac{-6}{5}$

Answer 1

Answer: (B)

$\frac{-5}{12}$

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

We know that $x^3+y^3+z^3-3xyz=(x+y+z{)}^3$

If $x+y+z=0$

Then $x^3+y^3+z^3-3xyz=0$

$\Rightarrow x^3+y^3+z^3=3xyz$

Here $a=\frac{1}{2},b=\frac{1}{3}c=\frac{-5}{6}$

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

So, ${\left(\frac{1}{2}\right)}^3+{\left(\frac{1}{3}\right)}^3-{\left(\frac{5}{6}\right)}^3=3\times \left(\frac{1}{2}\right)\times \left(\frac{1}{3}\right)\times (-\frac{5}{6})=-\frac{15}{36}=-\frac{5}{12}$