Simplify each of the following

Solution:

(a) $(x+3)^{3}+(x-3)^{3}$

The above equation is in the form of $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$

We know that, $a=(x+3), b=(x-3)$

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

$=(x+3+x-3)\left[(x+3)^{3}+(x-3)^{3}-(x+3)(x-3)\right]$

$=2 x\left[\left(x^{2}+3^{2}+2^{*} x^{*} 3\right)+\left(x^{2}+3^{2}-2^{*} x^{*} 3\right)-\left(x^{2}-3^{2}\right)\right]$

$=2 x\left[\left(x^{2}+9+6 x\right)+\left(x^{2}+9-6 x\right)-x^{2}+9\right]$

$=2 x\left[\left(x^{2}+9+6 x+x^{2}-9-6 x-x^{2}+9\right)\right]$

$=2 x\left(x^{2}+27\right)$

$=2 x^{3}+54 x$

Hence, the result of $(x+3)^{3}+(x-3)^{3}$ is $2 x^{3}+54 x$

(b) $(x / 2+y / 3)^{3}-(x / 2-y / 3)^{3}$

The above equation is in the form of $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$

We know that, $a=(x / 2+y / 3)^{3}, b=(x / 2-y / 3)^{3}$

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

$=\left[\left((x / 2+y / 3)^{3}-\left((x / 2-y / 3)^{3}\right)\right]\left[\left((x / 2+y / 3)^{3}\right)^{2}\left((x / 2-y / 3)^{3}\right)^{2}-\left((x / 2+y / 3)^{3}\right)\left((x / 2-y / 3)^{3}\right)\right.\right.$

$=(x / 3+y / 3-x / 2+y / 3)\left[\left((x / 2)^{2}+(y / 3)^{2}+(2 x y / 6)\right)+\left((x / 2)^{2}+(y / 3)^{2}-(2 x y / 6)\right)+\left((x / 2)^{2}-(y / 3)^{2}\right)\right]$

$=2 y / 3\left[\left(x^{2} / 4+y^{2} / 9+2 x y / 6\right)+\left(x^{2} / 4+y^{2} / 9-2 x y / 6\right)+x^{2} / 4-y^{2} / 9\right]$

$=2 y / 3\left[x^{2} / 4+y^{2} / 9+2 x y / 6+x^{2} / 4+y^{2} / 9-2 x y / 6+x^{2} / 4-y^{2} / 9\right]$

$=2 y / 3\left[x^{2} / 4+y^{2} / 9+x^{2} / 4+x^{2} / 4\right]$

$=2 y / 3\left[3 x^{2} / 4+y^{2} / 9\right]$

$=x^{2} y / 2+2 y^{3} / 27$

Hence, the result of $(x / 2+y / 3)^{3}-(x / 2-y / 3)^{3}=x^{2} y / 2+2 y^{3} / 27$

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

The above equation is in the form of $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$

We know that, $a=(x+2 x)^{3}, b=(x-2 x)^{3}$

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

$=(x+2 / x+x-2 / x)\left[(x+2 / x)^{2}+(x-2 / x)^{2}-((x+2 / x)(x-2 / x))\right]$

$=(2 x)\left[\left(x^{2}+4 / x^{2}+4 x / x\right)+\left(x^{2}+4 / x^{2}-4 x / x\right)-\left(x^{2}-4 / x^{2}\right)\right.$

$=(2 x)\left[\left(x^{2}+4 / x^{2}+4 x / x+x^{2}+4 / x^{2}-4 x / x-x^{2}+4 / x^{2}\right)\right.$

$=(2 x)\left[\left(x^{2}+4 / x^{2}+4 / x^{2}+4 / x^{2}\right)\right.$

$=(2 x)\left[\left(x^{2}+12 / x^{2}\right)\right.$

$=2 x^{3}+24 / x$

Hence, the result of $(x+2 / x)^{3}+(x-2 / x)^{3}=(2 x)\left[\left(x^{2}+12 / x^{2}\right)\right.$

(d) $(2 x-5 y)^{3}-(2 x+5 y)^{3}$

Given, $(2 x-5 y)^{3}-(2 x+5 y)^{3}$

The above equation is in the form of $a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)$

We know that, $a=(2 x-5 y), b=(2 x+5 y)$

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

$=(2 x-5 y-2 x-5 y)\left[(2 x-5 y)^{2}+(2 x+5 y)^{2}+\left((2 x-5 y)^{*}(2 x+5 y)\right)\right]$

$=(-10 y)\left[\left(4 x^{2}+25 y^{2}-20 x y\right)+\left(4 x^{2}+25 y^{2}+20 x y\right)+4 x^{2}-25 y^{2}\right]$

$=(-10 y)\left[4 x^{2}+25 y^{2}-20 x y+4 x^{2}+25 y^{2}+20 x y+4 x^{2}-25 y^{2}\right]$

$=(-10 y)\left[4 x^{2}+4 x^{2}+4 x^{2}+25 y^{2}\right]$

$=(-10 y)\left[12 x^{2}+25 y^{2}\right\}$

$=-120 x^{2} y-250 y^{3}$

Hence, the result of $(2 x-5 y)^{3}-(2 x+5 y)^{3}=-120 x^{2} y-250 y^{3}$