Without actually calculating the cubes,find t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $a = \frac{1}{2}$,$b = \frac{1}{3}$,and $c = -\frac{5}{6}$.First,calculate the sum $a + b + c$:$a + b + c = \frac{1}{2} + \frac{1}{3} - \frac{

Vedclass · Accepted Answer

$-\frac{5}{12}$

Vedclass · Answer

(C) Let $a = \frac{1}{2}$,$b = \frac{1}{3}$,and $c = -\frac{5}{6}$.First,calculate the sum $a + b + c$:$a + b + c = \frac{1}{2} + \frac{1}{3} - \frac{5}{6} = \frac{3 + 2 - 5}{6} = \frac{0}{6} = 0$.We know the algebraic identity: If $a + b + c = 0$,then $a^{3} + b^{3} + c^{3} = 3abc$.Applying this identity to the given expression:$\left(\frac{1}{2}ight)^{3} + \left(\frac{1}{3}ight)^{3} + \left(-\frac{5}{6}ight)^{3} = 3 	imes \left(\frac{1}{2}ight) 	imes \left(\frac{1}{3}ight) 	imes \left(-\frac{5}{6}ight)$.$= 3 	imes \frac{1}{6} 	imes \left(-\frac{5}{6}ight) = \frac{1}{2} 	imes \left(-\frac{5}{6}ight) = -\frac{5}{12}$.

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

Similar Questions

Evaluate $153 \times 147$.

Factorise the following expression: $27 x^{3}-8 y^{3}-54 x^{2} y+36 x y^{2}$

Evaluate the following product without multiplying directly: $76 \times 82$

Factorise $: 8 x^{3}+y^{3}-27 z^{3}+18 x y z$

Factorise the expression: $\frac{4x^2}{9} - \frac{x}{3} + \frac{1}{16}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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