Factorise : a³ - 8b³ - 64c³ - 24abc
Solution:
Given, the polynomial is a³ - 8b³ - 64c³ - 24abc ----- (1)
We have to factorise the polynomial.
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 = a³
x = a
y³ = -8b³
y = -2b
z³ = -64c³
z = -4c
Here x = a; y = -2b and z = -4c
So, (x + y + z) = (a - 2b - 4c)
(x² + y² + z²) = (a)² + (-2b)² + (-4c)²
= a² + 4b² + 16c²
(xy + yz + zx) = a(-2b) + (-2b)(-4c) + (-4c)a
= -2ab + 8bc - 4ac
Now, a³ - 8b³ - 64c³ - 24abc = (a - 2b - 4c)(a² + 4b² + 16c² - (-2ab + 8bc - 4ac))
= (a - 2b - 4c)(a² + 4b² + 16c² + 2ab - 8bc + 4ac)
Therefore, the factors are (a - 2b - 4c)(a² + 4b² + 16c² + 2ab - 8bc + 4ac)
✦ Try This: Factorise : 8a³ - 27b³ - 64c³ - 72abc
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 2
NCERT Exemplar Class 9 Maths Exercise 2.3 Problem 36(i)
Factorise : a³ - 8b³ - 64c³ - 24abc
Summary:
On factorising a³ - 8b³ - 64c³ - 24abc we get the factors as (a - 2b - 4c)(a² + 4b² + 16c² + 2ab - 8bc + 4ac
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