factorize 27a^3-8b^3+c^3+18abc
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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)
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