Question

factorise : 24a³+81b³​

Answer 1

Answer:

$3(2a + 3b)(4a^2 - 6ab + 9b^2)$

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 2

Answer:

24a³ + 81b³

= 3 ( 8a³ + 27b³ )

= 3 {(2a)³ + (3b)³}

= 3 ( 2a + 3b) {(2a)² - 2a × 3b + (3b)²}

= 3 ( 2a + 3b ) ( 4a² - 6ab + 9b²)

Hope it will help you.