Factorise : 24a³+81b³

1.	Factorise : 24a³+81b³
Answer» Answer: $<klux>3</klux>(2a + 3b)(4a^<klux>2</klux> - 6ab + 9b^2)$ Step-by-step explanation: Given Expression: $24a^3 + 81b^3$ From the given expression, take out 3 common. We will get $3(8a^3 + 27b^3)$ Now, LOOK at the expression in the bracket carefully. 8 is a cube of 2. [2³ = 8] a³ is a cube of a. [(a)³ = a³] 27 is a cube of 3 [3³ = 27] b³ is a cube of b [(b)³ = b³] Keeping these in mind, we can write the expression as $3[(2)^3a^3 + (3)^3b^3]$ This can further be WRITTEN as $3[(2a)^3 + (3b)^3]$ Now, the expression in the bracket is in the form of the identity $a^3 + b^3$ We know that $a^3 + b^3$ $= (a + b)(a^2 - ab + b^2)$ HENCE, we can write the expression as $3(2a + 3b)[(2a)^2 - (2a)(3b) + (3b)^2]$ Simplifying the expression, we finally get $3(2a + 3b)(4a^2 - 6ab + 9b^2)$

Answer»

Answer:

$<klux>3</klux>(2a + 3b)(4a^<klux>2</klux> - 6ab + 9b^2)$