Factorise using identity 27b^3-3b^5
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Step-by-step explanation:
8a
3
−27b
3
−64c
3
−72abc
=(2a)
3
+(−3b)
3
+(−4c)
3
−3×(2a)(−3b)(−4c)
Using the identity x
3
+y
3
+z
3
−3xyz=(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx)
where x=2a,y=−3b,z=−4c
=(2a−3b−4c)(4a
2
+9b
2
+16c
2
+6ab−12bc+8ac)
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