Question

Factorise each of the following
(i) 8a³ + b³+ 12a²b + 6ab²
(ii) 8a³- b³- 12a²b + 6ab²
(iii) 27 - 125a³- 135a + 225a²
(iv) 64a³-27b³ - 144a²b + 108ab²
(v) 27p³ - 1/216 - 9/2 p² + 1/4 p

Answer 1

$\mathbb{ ANSWER }$:

(i) 8a³ + b³+ 12a²b + 6ab²

It can be written as,

= (2a)³ + (b)³ + 3*(2a)(b) [2a+b]

= (2a + b)³

= (2a + b)(2a + b)(2a + b)


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(ii) 8a³- b³- 12a²b + 6ab²

It can be written as,

= (2a)³ - (b)³ - 3*(2a)(b) [2a - b]

= (2a - b)³

= (2a - b)(2a - b)(2a - b)


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(iii) 27 - 125a³- 135a + 225a²

It can be written as,

= (3)³ - (5a)³ - 3*(3)*(5a) [3 - 5a]

= (3 - 5a)³

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


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(iv) 64a³-27b³ - 144a²b + 108ab² 

It can be written as,

= (4a)³ - (3b)³ - 3*(4a)*(3b) [ 4a - 3b]

= (4a - 3b)³

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

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(v) 27p³ - 1/216 - 9/2 p² + 1/4 p

It can't be written as,

= (3p)³ - (1/6)³ - 3*(3p) (1/6) [ 3p - 1/6 ]

= (3p - 1/6)³

= (3p - 1/6) (3p - 1/6) (3p - 1/6)

Answer 2

Here is your solution

(i) 8a³ + b³+ 12a²b + 6ab²

=> (2a)³ + (b)³ + 3×(2a)(b) [2a+b]

=> (2a + b)³

=>(2a + b)(2a + b)(2a + b)

(ii) 8a³- b³- 12a²b + 6ab²

=> (2a)³ - (b)³ - 3*(2a)(b) (2a - b)

=> (2a - b)³

=> (2a - b)(2a - b)(2a - b)

(iii) 27 - 125a³- 135a + 225a²

=> (3)³ - (5a)³ - 3×(3)×(5a) (3 - 5a)

=> (3 - 5a)³

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

(iv) 64a³-27b³ - 144a²b + 108ab² 

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

=> (4a - 3b)³

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

(v) 27p³ - 1/216 - 9/2 p² + 1/4 p

=>(3p)³ - (1/6)³ - 3×(3p) (1/6) (3p - 1/6 )

=> (3p - 1/6)³

=>(3p - 1/6) (3p - 1/6) (3p - 1/6)

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