How do you factor $ 8{{x}^{3}}+1 $ ?

Hint: In the problem we have to factorize $ 8{{x}^{3}}+1 $ . So first we will factorize the coefficient of the $ {{x}^{3}} $ by checking the numbers which will divide the number $ 8 $ by giving zero reminder. Now we will use the exponential formula $ {{a}^{m}}{{b}^{m}}={{\left( ab \right)}^{m}} $ , we will simplify the term $ 8{{x}^{3}} $ . We know that $ {{1}^{n}}=1 $ , so we will use $ n=3 $ for this problem. Now the total equation is converted in the from of $ {{a}^{3}}+{{b}^{3}} $ . We have the formula $ {{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) $ and then we will get factors for the given equation.

Complete step by step answer:
Given equation $ 8{{x}^{3}}+1 $ .
In the above equation the coefficient of $ {{x}^{3}} $ is $ 8 $ . Factorizing the number $ 8 $ .Checking whether the number $ 8 $ is divisible by $ 2 $ or not. We can say that $ 8 $ is divisible by $ 2 $ and it will give $ 4 $ as a quotient and zero reminders. Now we can write the number $ 8 $ as
 $ 8=2\times 4 $
Factorizing the number $ 4 $ .
Checking whether the number $ 4 $ is divisible by $ 2 $ or not. We can say that $ 4 $ is divisible by $ 2 $
 And it will give $ 2 $ as quotient and zero reminder. Now we can write the number $ 4 $ as
 $ 4=2\times 2 $
From the above value we can write the number $ 8 $ as
 $ \begin{align}
  & 8=2\times 4 \\
 & \Rightarrow 8=2\times 2\times 2 \\
\end{align} $
We know that $ a.a.a.a.a...\text{ }n\text{ times}={{a}^{n}} $ , then
 $ 8={{2}^{3}} $
Substituting the above value in the given equation, then we can write
 $ 8{{x}^{3}}+1={{2}^{3}}{{x}^{3}}+1 $
Applying the formula $ {{a}^{m}}{{b}^{m}}={{\left( ab \right)}^{m}} $ , then we will get
 $ 8{{x}^{3}}+1={{\left( 2x \right)}^{3}}+1 $
We know that $ {{1}^{n}}=1 $ , let us assume $ n=3 $ for this problem, then $ {{1}^{3}}=1 $ . Substituting the above value in the above equation, then we will get
 $ 8{{x}^{3}}+1={{\left( 2x \right)}^{3}}+{{1}^{3}} $
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 have
 $ \begin{align}
  & 8{{x}^{3}}+1=\left( 2x+1 \right)\left( {{\left( 2x \right)}^{2}}-\left( 2x \right)\left( 1 \right)+{{1}^{2}} \right) \\
 & \therefore 8{{x}^{3}}+1=\left( 2x+1 \right)\left( 4{{x}^{2}}-2x+1 \right) \\
\end{align} $
Hence the factors of the given equation $ 8{{x}^{3}}+1 $ are $ 2x+1 $ , $ 4{{x}^{2}}-2x+1 $ .

Note:
 In the problem, we have the constant as $ 1 $ , so we used well know formula $ {{1}^{n}}=1 $ and solved the given equation. If the constant is other than $ 1 $ then we need to factorize the constant using the same method we have used to factorize the coefficient of $ {{x}^{3}} $ . Now we will use the algebraic formula to get the factors.