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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)\((x - \frac{2y}{3} )^3\)


Answer :

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

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