(i) (3x+4)3−(3x−4)3
=[(3x)3+(4)3+3(3x)(4)(3x+4)]−[(3x)2−(4)3−3(3x)(4)(3x−4)]
=[27x3+64+36x(3x+4)]−[27x3−64−36x(3x−4)]
=[27x3+64+108x2+144x]−[27x3−64−108x2+144x]
=27x3+64+108x2+144x−27x3+64+108x2−144x
=128+2162
(ii) (x+2x)3+(x−2x)3
=x3+(2x)3+3(x)(2x)(x+2x)+x3−(2x)3−3(x)(2x)(x−2x)
=x3+8x3+6x+12x+x3−8x3−6x+12x=2x3+24x