Without finding the cubes, factorise (x - y)³ + (y - z)³ + (z - x)³
Solution:
Given, (x - y)³ + (y - z)³ + (z - x)³
We have to factorize the polynomial without finding the cubes.
We know that a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)
If a + b + c = 0, then a³ + b³ + c³ - 3abc = 0 (or) a³ + b³ + c³ = 3abc.
Here, a = x - y
b = y - z
c = z - x
So, a + b + c = (x - y) + (y - z) + (z - x)
= x - y + y - z + z - x
= x - x + y - y + z - z
= 0
a + b + c = 0
So, a³ + b³ + c³ = 3abc
Therefore, (x - y)³ + (y - z)³ + (z - x)³ = 3(x - y)(y - z)(z - x)
✦ Try This: Without finding the cubes, factorise (2x - 3y)³ + (3y - 4z)³ + (4z - 2x)³
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 2
NCERT Exemplar Class 9 Maths Exercise 2.3 Sample Problem 3(ii)
Without finding the cubes, factorise (x - y)³ + (y - z)³ + (z - x)³
Summary:
Without finding the cubes, the value of (x - y)³ + (y - z)³ + (z - x)³ is 3(x - y)(y - z)(z - x)
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