FACTORISE

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

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2 thoughts on “FACTORISE<br /><br />1. 27-125a³-135a+225a²<br />2. 64a³-27b³-144a²b+108 ab²2<br /><br />❌DON’T SPAM ❌<br />CLASS 9 MATHS CHAPTER”

1. 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:

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

   =(3-5a)³

   Therefore,

   27-125a³-135a+225a = (3-5a)³

   Step-by-step explanation:

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2. 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)

   —————————————————————-

   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)

   –––––———————————————————–

   Used Identity:–

   • (a-b)^3 = a^3-3a^2b+3ab^2-b^3

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