How do you factor $64{x^3} + 8 = 0$?

Hint: To factorise a given cubic polynomial, we first assume the given polynomial to be $p\left( x \right)$. Then, make the coefficient of ${x^3}$ equal to one if it is not one and then find the constant term. Then, find all the factors of constant term. Then, check at which factor of constant term, $p\left( x \right)$ is zero by using trial method and get one factor of $p\left( x \right)$.Then, write $p\left( x \right)$ as the product of this factor and a quadratic polynomial. This will be the desired factor of the given polynomial.

Complete step by step solution:
Given cubic polynomial: $64{x^3} + 8$
We have to factorize this cubic polynomial.
So, assuming a given polynomial to be $p\left( x \right)$.
Let $p\left( x \right) = 64{x^3} + 8$
Next step is to make the coefficient of ${x^3}$ equal to one if it is not one.
So, $p\left( x \right) = 64\left( {{x^3} + \dfrac{1}{8}} \right)$

Next step is to find the constant term and all the factors of the constant term.
So, we have to find the constant term of $p\left( x \right)$ and all its factors.
Constant term is $\dfrac{1}{8}$ and all factors of $\dfrac{1}{8}$ are \[ \pm 1, \pm \dfrac{1}{2}, \pm \dfrac{1}{4}, \pm \dfrac{1}{8}\].

Next step is to check at which factor of constant term, $p\left( x \right)$ is zero by using trial method and get one factor of $p\left( x \right)$.
So, put $x = - \dfrac{1}{2}$ in $p\left( x \right) = 64\left( {{x^3} + \dfrac{1}{8}} \right)$.
$p\left( { - \dfrac{1}{2}} \right) = 64\left[ {{{\left( { - \dfrac{1}{2}} \right)}^3} + \dfrac{1}{8}} \right]$
$ \Rightarrow p\left( { - \dfrac{1}{2}} \right) = 64\left[ { - \dfrac{1}{8} + \dfrac{1}{8}} \right]$
$ \Rightarrow p\left( { - \dfrac{1}{2}} \right) = 0$
So, $\left( {x + \dfrac{1}{2}} \right)$ is a factor of $p\left( x \right)$.
Next step is to write $p\left( x \right)$ as the product of this factor and a quadratic polynomial.
So, divide $p\left( x \right)$ by $\left( {x + \dfrac{1}{2}} \right)$.

$ \Rightarrow p\left( x \right) = 64\left( {x + \dfrac{1}{2}} \right)\left( {{x^2} - \dfrac{1}{2}x + \dfrac{1}{4}} \right)$
$ \Rightarrow 64{x^3} + 8 = 8\left( {2x + 1} \right)\left( {4{x^2} - 2x + 1} \right)$

Therefore, $64{x^3} + 8 = 0$ can be factored as $8\left( {2x + 1} \right)\left( {4{x^2} - 2x + 1} \right) = 0$.

Note: We can also factorise given cubic polynomial using algebraic identity ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$.
Since, $64{x^3}$ can be written as ${\left( {4x} \right)^3}$ and $8$ can be written as ${2^3}$.
So, both terms are perfect cubes in $64{x^3} + 8$. Thus, factor using the algebraic identity ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$ by substituting $a = 4x$ and $b = 2$.
$ \Rightarrow 64{x^3} + 8 = \left( {4x + 2} \right)\left( {16{x^2} - 8x + 4} \right)$
$ \Rightarrow 64{x^3} + 8 = 8\left( {2x + 1} \right)\left( {4{x^2} - 2x + 1} \right)$