Factorise each of the following: 27y3+125z3 [ 3 MARKS]
Formula: 1 Mark Application: 2 Marks 27y3+125z3
=(3y)3+(5z)3
Using the identity x3+y3=(x+y)(x2+y2−xy) =(3y+5z)((3y)2+(5z)2−(3y)(5z))
=(3y+5z)(9y2+25z2−15yz)