Question

factorise 27a3+8b3+c3-18abc​

Answer 1

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)