Factorise : 2√2a³ + 8b³ - 27c³ + 18√2abc
Solution:
Given, the polynomial is 2√2a³ + 8b³ - 27c³ + 18√2abc
We have to factorise the polynomial.
The expression can be written as (√2a)³ + (2b)³ - (3c)³ + 18√2abc --- (1)
Using the algebraic identity,
x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx) --- (2)
x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - (xy + yz + zx))
On comparing (1) and (2),
x³ = (√2a)³
x = √2a
y³ = (2b)³
y = 2b
z³ = -(3c)³
z = -3c
Here x = √2a; y = 2b and z = -3c
So, (x + y + z) = (√2a + 2b - 3c)
(x² + y² + z²) = (√2a)² + (2b)² + (-3c)²
= 2a² + 4b² + 9c²
(xy + yz + zx) = √2a(2b) + (2b)(-3c) + (-3c)√2a
= 2√2ab - 6bc - 3√2ac
Now, 2√2a³ - 8b³ - 64c³ - 24abc = (√2a + 2b - 3c)(2a² + 4b² + 9c² - (2√2ab - 6bc - 3√2ac))
= (√2a + 2b - 3c)(2a² + 4b² + 9c² - 2√2ab + 6bc + 3√2ac)
Therefore, the factors are (√2a + 2b - 3c)(2a² + 4b² + 9c² - 2√2ab + 6bc + 3√2ac)
✦ Try This: Factorise : 8a³ - b³ - 125c³ - 30abc
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 2
NCERT Exemplar Class 9 Maths Exercise 2.3 Problem 36(ii)
Factorise : 2√2a³ + 8b³ - 27c³ + 18√2abc
Summary:
On factorising 2√2a³ + 8b³ - 27c³ + 18√2abc we get the factors as (√2a + 2b - 3c)(2a² + 4b² + 9c² - 2√2ab + 6bc + 3√2ac)
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