How do you factor $16{{x}^{3}}-250$?

Hint: We can observe that the given equation contains one cubic variable in subtraction. In algebra we have the formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$. But in the given equation we have additionally coefficient for the variable and a constant, so we need to factorize the coefficient of the given variable, constant at the same time and we will convert the given equation in form of ${{a}^{3}}+{{b}^{3}}$, by applying some exponential rules, then we will use the algebraic formula and write the given equation as the product of its factors.

Complete step-by-step solution:
Given equation $16{{x}^{3}}-250$.
Considering the coefficient of ${{x}^{3}}$, which is $16$. Factoring $16$.
Dividing $16$ with $2$. We will get $8$ as quotient, so we can write $16$ as
$16=2\times 8$.
Now dividing $8$ with $2$. We will get $4$ as quotient, so we can write $8$ as
$8=2\times 4$.
Now dividing $4$ with $2$. We will get $2$ as quotient, so we can write $4$ as
$4=2\times 2$.
From the all the above values we can write the number $16$ as
$\begin{align}
  & 16=2\times 8 \\
 & \Rightarrow 16=2\times 2\times 4 \\
 & \Rightarrow 16=2\times 2\times 2\times 2 \\
\end{align}$
We have an exponential rule $a\times a\times a\times ....n\text{ times}={{a}^{n}}$, then we will have
$\therefore 16={{2}^{4}}$
Considering the constant which is $250$. Factoring $250$.
Dividing $250$ with $2$. We will get $125$ as quotient, so we can write
$250=2\times 125$
We can write the value $125$ as $125=5\times 5\times 5$. From the exponential rule $a\times a\times a\times ....n\text{ times}={{a}^{n}}$, we can write $125$ as $125={{5}^{3}}$
Now the value $250$ will be written as
$250=2\times {{5}^{3}}$.
Now the given equation is modified as
$16{{x}^{3}}-250={{2}^{4}}{{x}^{3}}-2\times {{5}^{3}}$
Taking $2$ as common from the above equation, then we will have
$\Rightarrow 16{{x}^{3}}-250=2\left( {{2}^{3}}{{x}^{3}}-{{5}^{3}} \right)$
We have the exponential rule ${{a}^{m}}{{b}^{m}}={{\left( ab \right)}^{m}}$, then we will get
$\Rightarrow 16{{x}^{3}}-250=2\left[ {{\left( 2x \right)}^{3}}-{{5}^{3}} \right]$
Applying the formula ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$ in the above equation, then we will get
$\begin{align}
  & \Rightarrow 16{{x}^{3}}-250=2\left[ \left( 2x-5 \right)\left( {{\left( 2x \right)}^{2}}+\left( 2x \right)\left( 5 \right)+{{5}^{2}} \right) \right] \\
 & \Rightarrow 16{{x}^{3}}-250=2\left( 2x-5 \right)\left( 4{{x}^{2}}+10x+25 \right) \\
\end{align}$

Note: When we multiply the above obtained factors, we need to get the given equation as result. If you don’t get the given equation as a result, then our solution is not correct. If they are the same then our result is correct.