Question

Without actually calculating the cubes, find the value of (1/2)3 + (1/3)3 + (-5/6)3​

Answer 1

Answer:

Step-by-step explanation:

according to the identity

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

here

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

⇒$\frac{1}{8} + \frac{1}{27} +\frac{-125}{216} - 3(\frac{1}{2} )(\frac{1}{3} )(\frac{-5}{6} )$ = $(\frac{1}{2} +\frac{1}{3}+\frac{-5}{6} )(\frac{1}{2} ^{2} +\frac{1}{3} ^{2} +\frac{-5}{6} ^{2})$