# Find the cube of each of the following binomial expression

Question:

Find the cube of each of the following binomial expression

(a) $(1 / x+y / 3)$

(b) $\left(3 / x-2 / x^{2}\right)$

(c) $(2 x+3 / x)$

(d) $(4-1 / 3 x)$

Solution:

(a)  Given,

$(1 / x+y / 3))^{3}$

The above equation is in the form of $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$

We know that, $a=1 / x, b=y / 3$

By using $(a+b)^{3}$ formula

$(1 / x+y / 3))^{3}$

$=(1 / x)^{3}+(y / 3)^{3}+3(1 / x)(y / 3)(1 / x+y / 3)$

$=1 / x^{3}+y^{3} / 27+3 * 1 / x^{*} y / 3(1 / x+y / 3)$

$=1 / x^{3}+y^{3} / 27+y / x(1 / x+y / 3)$

$=1 / x^{3}+y^{3} / 27+\left(y / x^{*} 1 / x\right)+\left(y / x^{*} y^{3}\right)$

$\left.=1 / x^{3}+y^{3} / 27+y / x^{2}+y^{2} / 3 x\right)$

Hence,

$\left.(1 / x+y / 3))^{3}=1 / x^{3}+y^{3} / 27+y / x^{2}+y^{2} / 3 x\right)$

(b) Given,

$\left(\left(3 / x-2 / x^{2}\right)\right)^{3}$

The above equation is in the form of $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$

We know that, $a=3 / x, b=2 / x^{2}$

By using $(a-b)^{3}$ formula

$\left(\left(3 / x-2 / x^{2}\right)\right)^{3}=(3 x)^{3}-\left(2 / x^{2}\right)^{3}-3(3 / x)\left(2 / x^{2}\right)\left(3 / x-2 / x^{2}\right)$

$=27 / x^{3}-8 / x^{6}-3^{2} 3 / x * 2 / x^{2}\left(3 / x-2 / x^{2}\right)$

$=27 / x^{3}-8 / x^{6}-18 / x^{3}\left(3 / x-2 / x^{2}\right)$

$=27 / x^{3}-8 / x^{6}-\left(18 / x^{3} * 3 / x\right)+\left(18 / x^{3} * 2 / x^{2}\right)$

$=27 / x^{3}-8 / x^{6}-54 / x^{4}+36 / x^{5}$

Hence, $\left(\left(3 / x-2 / x^{2}\right)\right)^{3}=27 / x^{3}-8 / x^{6}-54 / x^{4}+36 / x^{5}$

(c) Given,

$(2 x+3 / x)^{3}$

The above equation is in the form of $(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)$

We know that, $a=2 x, b=3 / x$

By using $(a+b)^{3}$ formula

$=8 x^{3}+27 / x^{3}+18 x / x(2 / x+3 / x)$

$=8 x^{3}+27 / x^{3}+18 x / x(2 x+3 / x)$

$=8 x^{3}+27 / x^{3}+\left(18^{*} 2 x\right)+\left(18^{*} 3 / x\right)$

$\left.=8 x^{3}+27 / x^{3}+36 \times 54 / x\right)$

Hence,

The cube of $\left.(2 x+3 / x)^{3}=8 x^{3}+27 / x^{3}+36 \times 54 / x\right)$

(d) Given,

$(4-1 / 3 x)^{3}$

The above equation is in the form of $(a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)$

We know that, $a=4, b=1 / 3 x$

By using $(a-b)^{3}$ formula

$(4-1 / 3 x)^{3}=4^{3}-(1 / 3 x)^{3}-3(4)(1 / 3 x)(4-1 / 3 x)$

$=64-1 / 27 x^{3}-12 / 3 x(4-1 / 3 x)$

$=64-1 / 27 x^{3}-4 / x(4-1 / 3 x)$

$=64-1 / 27 x^{3}-(4 / 3 x * 4)+(4 / 3 x * 1 / 3 x)$

$=64-1 / 27 x^{3}-16 / x+\left(4 / 3 x^{2}\right)$

Hence,

The cube of $(4-1 / 3 x)^{3}=64-1 / 27 x^{3}-16 / x+\left(4 / 3 x^{2}\right)$