Question

Without calculating cubes find value of (x-2y)³+(2y-3z)³+(3z-x)³

Answer 1

Answer:

• it's value is

$3(x - 2y) (2y - 3z)(3z - x)$

• since ...

a^3+b^3+c^3=3abc when (a+b+c)=0

• now here,

(x-2y)+(2y-3z)+(3z-x)=0

• so,

(x-2y)³+(2y-3z)³+(3z-x)³

=3(x - 2y) (2y - 3z)(3z - x)

Answer 2

Answer:

Let, x-2y= a , 2y-3z = b and 3z-x = c

Clearly, a+b+c= 0

Therefore,

(x-2y)^3 + (2y-3z)^3 + (3z-x)^3 = 3(x-2y)(2y-3z)(3z-x)

Step-by-step explanation:

Since, by the basic algebraic identity

if , a+b+c = 0

a^3 + b^3 + c^3 = 3abc

Hope this helps you !