Question

FACTORISE

1. 27-125a³-135a+225a²
2. 64a³-27b³-144a²b+108 ab²2

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CLASS 9 MATHS CHAPTER 2 POLYNOMIALS​

Answer 1

Step-by-step explanation:

Given :-

1. 27-125a^3-135a+225a^2

2. 64a^3-27b^3-144a^2b+108 ab^2

To find:-

Factorise the following expressions ?

Solution:-

1)

Given expression is 27-125a^3-135a+225a^2

It can be written as

=>(3)^3-5^3a^3-3(9)(5a)+3(3)(25)a^2

=> (3)^3-5^3a^3-3(3^2)(5a)+3(3)(5a)^2

=> (3)^3-3(3^2)(5a)+3(3)(5a)^2-(5a)^3

This is in the form of a^3-3a^2b+3ab^2-b^3

Where a = 3 and b = 5a

We know that

(a-b)^3 = a^3-3a^2b+3ab^2-b^3

=> (3)^3-3(3^2)(5a)+3(3)(5a)^2-(5a)^3

=> (3-5a)^3

=> (3-5a)(3-5a)(3-5a)

27-125a^3-135a+225a^2

= (3-5a)(3-5a)(3-5a)

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2)

Given expression is 64a^3-27b^3-144a^2b+108 ab^2

It can be written as

=>4^3a^3-3^3b^3-3(16a^2)(3b)+3(4a)(9b^2)

=>( 4a)^3-(3b)^3-3(4a)^2(3b)+3(4a)(3b)^2

=> ( 4a)^3-3(4a)^2(3b)+3(4a)(3b)^2-(3b)^3

This is in the form of a^3-3a^2b+3ab^2-b^3

Where a = 4a and b = 3b

We know that

(a-b)^3 = a^3-3a^2b+3ab^2-b^3

=>( 4a)^3-3(4a)^2(3b)+3(4a)(3b)^2-(3b)^3

=> (4a-3b)^3

=> (4a-3b)(4a-3b)(4a-3b)

64a^3-27b^3-144a^2b+108 ab^2 = (4a-3b)(4a-3b)(4a-3b)

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Used Identity:-

• (a-b)^3 = a^3-3a^2b+3ab^2-b^3

Answer 2

Answer:

27-125a³-135a+225a

= (3-5a)³

= (3-5a)(3-5a)(3-5a)

Given 27-125a³-135a+225a

=3³+(-5a)³+3×3²×(-5a)+3×3×(-5a)²

= [3+(-5a)]³

By an algebraic identity:

$\boxed {x^{3}+y^{3}+3x^{2}y+3xy^{2}\\=(x+y)^{3}} */$

=(3-5a)³

Therefore,  

27-125a³-135a+225a = (3-5a)³

Step-by-step explanation:

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