Question

Factorise
i) 27-125a^3-135a+225a^2
ii)64a^3-27b^3-144a^2b+108ab^2
iii)27p^3-1/216-9/2p+1/4p

Answer 1

1) 27 − 125a3 − 135a + 225a2=(3)3 − (5a)3 − 3(3)2(5a) + 3(3)(5a)2=(3−5a)3=(3−5a)(3−5a)(3−5a)
2) 64a3−27b3−144a2b+108ab2=(4a)3−(3b)3−3(4a)2(3b)+3(4a)(3b)2=(4a−3b)3 [(a−b)3=a3−b3−3a2b+3ab2]
Regards
3) I don't know

Answer 2

iii) 27p^3-1/216-9/2p^2+1/4       (3p)^3-(1/6)^3-(3p)^2/2+(1/2)^2p (3p-1/6)^3 (3p-1/6)(3p-1/6)(3p-1/6) We will open the cube.