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