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# Write the following cubes in expanded form.i)$$(2x + 1)^3$$ii)$$(2a - 3b)^3$$iii)$$[\frac{3x}{2} + 1]^3$$iv)$$(x - \frac{2y}{3} )^3$$

i)$$(2x + 1)^3$$
$$= (2x)^3 + (1)^3 + 3(2x)(1)(2x + 1)$$
$$[\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)]$$
$$= 8x^3 + 1 + 6x(2x + 1))$$
$$= 8x^3 + 1 + 12x^2 + 6x$$
$$= 8x^3 + 12x^2 + 6x + 1$$

ii)$$(2a - 3b)^3$$
$$= (2a)^3 - (3b)^3 - 3(2a)(3b)(2a - 3b)$$
$$[\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)]$$
$$= 8a^3 - 27b^3 - 18ab(2a - 3b))$$
$$= 8a^3 - 27b^3 - 36a^2b + 54ab^2)$$
$$= 8a^3 - 36a^2b + 54ab^2 - 27b^3$$

iii)$$[\frac{3x}{2} + 1]^3$$
$$= (\frac{3x}{2} )^3 + (1)^3 + 3(\frac{3x}{2} )(1)(\frac{3x}{2} + 1)$$
$$[\because (a + b)^3 = a^3 + b^3 + 3ab(a + b)]$$
$$= \frac{27x^3}{8} + 1 + \frac{9x}{2} (\frac{3x}{2} + 1))$$
$$= \frac{27x^3}{8} + 1 + \frac{27x^2}{4} + \frac{9x}{2}$$
$$= \frac{27x^3}{8} + \frac{27x^2}{4} + \frac{9x}{2} + 1$$

iv)$$(x - \frac{2y}{3} )^3 = (x)^3 - (\frac{2y}{3} )^3 - 3(x)(\frac{2y}{3} )(x - \frac{2y}{3} )$$
$$[\because (a - b)^3 = a^3 - b^3 - 3ab(a - b)]$$
$$= x^3 - \frac{8y^3}{27} - 2xy(x - \frac{2y}{3} )$$
$$= x^3 - \frac{8y^3}{27} - 2x^2y + \frac{4xy^2}{3}$$
$$= x^3 - 2x^2y + \frac{4xy^2}{3} - \frac{8y^3}{27}$$