Write the following cubes in expanded form:

(i) $(2 x+1)^{3}$

(ii) $(2 a-3 b)^{3}$

(iii) $\left[\frac{3}{2} x+1\right]^{3}$

(iv) $\left[x-\frac{2}{3} y\right]^{3}$
Solution:

It is known that,

$(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$

and $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$

(i) $(2 x+1)^{3}=(2 x)^{3}+(1)^{3}+3(2 x)(1)(2 x+1)$

$=8 x^{3}+1+6 x(2 x+1)$

$=8 x^{3}+1+12 x^{2}+6 x$

$=8 x^{3}+12 x^{2}+6 x+1$

(ii) $(2 a-3 b)^{3}=(2 a)^{3}-(3 b)^{3}-3(2 a)(3 b)(2 a-3 b)$

$=8 a^{3}-27 b^{3}-18 a b(2 a-3 b)$

$=8 a^{3}-27 b^{3}-36 a^{2} b+54 a b^{2}$

(iii) $\left[\frac{3}{2} x+1\right]^{3}=\left[\frac{3}{2} x\right]^{3}+(1)^{3}+3\left(\frac{3}{2} x\right)(1)\left(\frac{3}{2} x+1\right)$

$=\frac{27}{8} x^{3}+1+\frac{9}{2} x\left(\frac{3}{2} x+1\right)$

$=\frac{27}{8} x^{3}+1+\frac{27}{4} x^{2}+\frac{9}{2} x$

$=\frac{27}{8} x^{3}+\frac{27}{4} x^{2}+\frac{9}{2} x+1$

(vi) $\left[x-\frac{2}{3} y\right]^{3}=x^{3}-\left(\frac{2}{3} y\right)^{3}-3(x)\left(\frac{2}{3} y\right)\left(x-\frac{2}{3} y\right)$

$=x^{3}-\frac{8}{27} y^{3}-2 x y\left(x-\frac{2}{3} y\right)$

$=x^{3}-\frac{8}{27} y^{3}-2 x^{2} y+\frac{4}{3} x y^{2}$
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