How do you factor $64{{x}^{3}}-1$?

Hint: To solve the given expression first we will convert it into ${{a}^{3}}-{{b}^{3}}$ form and then we will apply the formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ to find the factors. Then we will use a quadratic formula to factorize the obtained equation further. The quadratic formula is given as
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

Complete step by step answer:
We have been given an expression $64{{x}^{3}}-1$.
We have to find the factors of the given expression.
Now, we know that $64={{4}^{3}}$ so we can write the given equation as
$\Rightarrow {{\left( 4x \right)}^{3}}-{{1}^{3}}$
Now, we know that difference of cubes formula is given as ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$
Now, by applying the formula in the above obtained equation we will get
$\Rightarrow {{\left( 4x \right)}^{3}}-{{1}^{3}}=\left( 4x-1 \right)\left( {{\left( 4x \right)}^{2}}+4x+{{1}^{2}} \right)$
Now, simplifying the above obtained equation we will get
$\Rightarrow {{\left( 4x \right)}^{3}}-{{1}^{3}}=\left( 4x-1 \right)\left( 16{{x}^{2}}+4x+1 \right)$
Now we will apply the quadratic formula to factorize $\left( 16{{x}^{2}}+4x+1 \right)$. Then we will get
$\Rightarrow x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Now, substituting the values and solving further we will get
\[\begin{align}
  & \Rightarrow x=\dfrac{-4\pm \sqrt{{{4}^{2}}-4\times 16\times 1}}{2\times 16} \\
 & \Rightarrow x=\dfrac{-4\pm \sqrt{16-64}}{32} \\
 & \Rightarrow x=\dfrac{-4\pm \sqrt{-48}}{32} \\
 & \Rightarrow x=\dfrac{-4\pm 4\sqrt{-3}}{32} \\
\end{align}\]
Now, taking common terms out and solving further we will get
$\begin{align}
  & \Rightarrow x=\dfrac{4\left( -1\pm 1\sqrt{-3} \right)}{32} \\
 & \Rightarrow x=\dfrac{\left( -1\pm 1\sqrt{-3} \right)}{8} \\
\end{align}$
Now, we know that $\sqrt{-1}=i$ so we get two roots of the above obtained equation as $\Rightarrow x=\dfrac{\left( -1+1\sqrt{3}i \right)}{8},\dfrac{\left( -1-1\sqrt{3}i \right)}{8}$

So we get the factors of the equation $64{{x}^{3}}-1$ as $\left( 4x-1 \right),\left( 4x+\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i \right),\left( 4x+\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i \right)$

Note: Here in this question we factorize the quadratic equation by using the quadratic formula. We can also use other methods like splitting the middle term, factorization method to solve the equation. Students must remember the value of $\sqrt{-1}=i$ to factorize further. As $i$ is the imaginary part in the complex numbers, so the obtained roots are complex numbers.