Question

without actually calculating the cubes find the value of (1/2x-1/3y)3 +(1/3y-1/4z)3 + (1/4z-1/2x)3.​

Answer 1

Step-by-step explanation:

There's a formula that

$a {}^{3} + b {}^{3} + c {}^{3} = 3abc$

if a=b=c or a+b+c =0

Here, if you add all the terms inside the bracket, you will get zero, so applying this formula

$\\ = 3( \frac{1 }{2x} - \frac{1}{3y} )( \frac{1}{3y} - \frac{1}{4z} )( \frac{1}{4z} - \frac{1}{2x} )$

is the answer