Question

Factorise 27a3 +b3+8c3 - 18abc using Indentity?​

Answer 1

Answer:

(3a + b + 2c) ( 9a² + b² + 4c² -3ab - 2bc - 6ac)

Explanation:-

Here we have,

27+ b³ + 8c³ - 18abc

= (3a)+ (b)³ + (2c)³ - 3(3a)(b)(2c)

= ( 3a + b + 2c) [ (3a)² + b² + (2c)² - (3a)(b) - (b)(2c)-(2c)(3a)

= (3a + b + 2c) ( 9a² + b² + 4c² -3ab - 2bc - 6ac)

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Answer 2

Answer:

We have 27a

3

+b

3

+8c

3

−18abc =(3a)

3

+(b)

3

+(2c)

3

−3(3a)(b)(2c)

We know that x

3

+y

3

+z

3

−3xyz=(x+y+z)(x

2

+y

2

+z

2

−xy−yz−zx)

Comparing both, we get

x=3a,y=b and z=2c

Then (3a+b+2c)[(3a)

2

+(b)

2

+(2c)

2

−(3a)(b)−(b)(2c)−(2c)(3a)]

=(3a+b+2c)(9a

2

+b

2

+4c

2

−3ab−2bc−6ac)