Write the following cubes in expanded form: (i) (2x+1)³ (ii) (2a-b)³ (iii) (3x+1)³ (iv) (x-3y)³

Since, we have
(a+b)³=a³+b³+3ab(a+b) (i)
(a-b)³=a³-b³-3ab(a-b). (ii)
(i)Given (2x+1)³=(2x)³+(1)³+3(2x)(1)(2x+1) [Substitute a=2x and b=1 in equation (i)]
=8x³+1+6x(2x+1)
=8x³+12x²+6x+1
(ii)(2a-b)³=(2a)³-(b)³+3(2a)(b)(2a-b) [ from equation(ii)]
=8a³-b³-6a²b+6ab²
(iii)(3x+1)³=(3x)³+(1)³+3(3x)(1)(3x+1) [from equation(i)]
=27x³+1+9x(3x+1)
=27x³+27x²+9x+1
(iv)(x-3y)³=x³-(3y)³-3x(3y)(x-3y) [From equation(ii)]
=x³-(27)y³-9x²(y)+27xy²
=x³-27y³-9x²y+27xy²