Without finding the cubes, factorise : (x - 2y)³ + (2y - 3z)³ + (3z - x)³
Solution:
Given, the polynomial is (x - 2y)³ + (2y - 3z)³ + (3z - x)³
We have to factorise the polynomial without finding the cubes.
Using the algebraic identity,
a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)
If a + b + c = 0, then a³ + b³ + c³ - 3abc = 0
So, a³ + b³ +c³ = 3abc.
Here, a = x - 2y; b = 2y - 3z; c = 3z - x
a + b + c = x - 2y + 2y - 3z + 3z - x
= x - x + 2y - 2y + 3z - 3z
= 0
a + b + c = 0
Hence, a³ + b³ + c³ = 3abc.
3abc = 3(x - 2y)(2y - 3z)(3z - x)
Therefore, the factors are 3(x - 2y)(2y - 3z)(3z - 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 Problem 38
Without finding the cubes, factorise : (x - 2y)³ + (2y - 3z)³ + (3z - x)³
Summary:
On factorising (x - 2y)³ + (2y - 3z)³ + (3z - x)³ without finding the cubes the factors are 3(x - 2y)(2y - 3z)(3z - x)
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