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

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To do:

We have to write the given cubes in expanded form.

Solution:

We know that,

$(a+b)^3=a^3+b^3+3ab(a+b)$

$(a-b)^3=a^3-b^3-3ab(a-b)$

Therefore,

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

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

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

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

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

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

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

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

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

Hence $(2 a-3 b)^{3}=8a^3 - 36a^2b + 54ab^2 - 27 b^3$

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

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

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

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

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

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

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

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

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

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

Updated on 10-Oct-2022 13:39:07