Factorize $8x^3-\dfrac{1}{27y^3}$

Now, let’s try to express each term in cube form and it is possible when the numbers $8$ and $27$ are written in cube form. The number $2$ is a factor of $8$, the number $3$ is a factor of $27$ and the cubes of numbers $2$ and $3$ are $8$ and $27$. So, let’s express the numbers $8$ and $27$ in cube form.

$=\,\,$ $2 \times 2 \times 2 \times x^3$ $-$ $\dfrac{1}{3 \times 3 \times 3 \times y^3}$

$=\,\,$ $2^3 \times x^3$ $-$ $\dfrac{1}{3^3 \times y^3}$

The product of the quantities in cube form in both terms can be evaluated by the power rule of a product.

$=\,\,$ $(2 \times x)^3$ $-$ $\dfrac{1}{(3 \times y)^3}$

$=\,\,$ $(2x)^3$ $-$ $\dfrac{1}{(3y)^3}$

According to the properties of factors, the number one is a factor of itself. So, the number $1$ can be written as follows.

$=\,\,$ $(2x)^3$ $-$ $\dfrac{1 \times 1 \times 1}{(3y)^3}$

$=\,\,$ $(2x)^3$ $-$ $\dfrac{1^3}{(3y)^3}$

The power rule of quotient can be used to find the quotient of the cubes.

$=\,\,$ $(2x)^3$ $-$ $\Big(\dfrac{1}{3y}\Big)^3$

Now, let’s use the difference of two cubes formula to factor the simplified polynomial.

$a^3-b^3$ $\,=\,$ $(a-b)(a^2+ab+b^2)$

Let’s take $a$ $\,=\,$ $2x$ and $b$ $\,=\,$ $\dfrac{1}{3y}$. Substitute them in the difference of cubes formula to factorise the above algebraic expression.

$=\,\,$ $\bigg(2x-\dfrac{1}{3y}\bigg)\Bigg((2x)^2+2x \times \dfrac{1}{3y}+\bigg(\dfrac{1}{3y}\bigg)^2\Bigg)$

The first factor of the expression is already in simplified form. Now, let’s focus on simplifying the second factor of the expression.

$=\,\,$ $\bigg(2x-\dfrac{1}{3y}\bigg)\Bigg(2x \times 2x+2x \times \dfrac{1}{3y}+\bigg(\dfrac{1}{3y}\bigg)^2\Bigg)$

$=\,\,$ $\bigg(2x-\dfrac{1}{3y}\bigg)\Bigg(4x^2+2x \times \dfrac{1}{3y}+\bigg(\dfrac{1}{3y}\bigg)^2\Bigg)$

The first factor is $2x$. The second factor of the term is the reciprocal of $3y$ and it is a fraction. So, let’s express $2x$ in fraction form.

$=\,\,$ $\bigg(2x-\dfrac{1}{3y}\bigg)\Bigg(4x^2+\dfrac{2x}{1} \times \dfrac{1}{3y}+\bigg(\dfrac{1}{3y}\bigg)^2\Bigg)$

Now, multiply the fractions to find the product of them and it can be done by the multiplication of the fractions.

$=\,\,$ $\bigg(2x-\dfrac{1}{3y}\bigg)\Bigg(4x^2+\dfrac{2x \times 1}{1 \times 3y}+\bigg(\dfrac{1}{3y}\bigg)^2\Bigg)$

$=\,\,$ $\bigg(2x-\dfrac{1}{3y}\bigg)\Bigg(4x^2+\dfrac{2x}{3y}+\bigg(\dfrac{1}{3y}\bigg)^2\Bigg)$

Finally, let’s find the square of the reciprocal of $3y$ to express the factor form of the given algebraic expression.

$=\,\,$ $\bigg(2x-\dfrac{1}{3y}\bigg)\Bigg(4x^2+\dfrac{2x}{3y}+\dfrac{1}{3y} \times \dfrac{1}{3y}\Bigg)$

Now, use the multiplication of the fractions to multiply the reciprocal of $3y$ by itself.

$=\,\,$ $\bigg(2x-\dfrac{1}{3y}\bigg)\Bigg(4x^2+\dfrac{2x}{3y}+\dfrac{1 \times 1}{3y \times 3y}\Bigg)$

$=\,\,$ $\bigg(2x-\dfrac{1}{3y}\bigg)\Bigg(4x^2+\dfrac{2x}{3y}+\dfrac{1}{9y^2}\Bigg)$