How do you factor completely: $81{{x}^{4}}-16$

Hint: To factor the given algebraic expression i.e. $81{{x}^{4}}-16$, we can write $81{{x}^{4}}$ as ${{\left( 3x \right)}^{4}}$ and 16 as ${{2}^{4}}$ then we will require this algebraic property in which ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. But you might think we have shown the power of 4 then how can we reduce this fourth power to second power. We can achieve this by writing ${{\left( 3x \right)}^{4}}$ as ${{\left( {{\left( 3x \right)}^{2}} \right)}^{2}}$ and ${{2}^{4}}$ as ${{\left( {{2}^{2}} \right)}^{2}}$.

Complete step by step answer:
The algebraic expression given in the above problem which we are asked to factorize is as follows:
$81{{x}^{4}}-16$
Now, we know that 81 is fourth power of 3 and 16 is fourth power of 2 so using this power relations along with that we can write ${{x}^{4}}$ with 81 as ${{\left( 3x \right)}^{4}}$ in the above we get,
$\Rightarrow {{\left( 3x \right)}^{4}}-{{\left( 2 \right)}^{4}}$
We know the algebraic identity which states that:
${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
We can apply the above identity in ${{\left( 3x \right)}^{4}}-{{\left( 2 \right)}^{4}}$ by first rearranging this expression in the form of $\left( {{a}^{2}}-{{b}^{2}} \right)$ we get,
$\Rightarrow {{\left( {{\left( 3x \right)}^{2}} \right)}^{2}}-{{\left( {{2}^{2}} \right)}^{2}}$
Now, applying $\left( {{a}^{2}}-{{b}^{2}} \right)$ identity in the above then $a={{\left( 3x \right)}^{2}}$ and $b={{2}^{2}}$ we get,
$\Rightarrow \left( {{\left( 3x \right)}^{2}}+{{2}^{2}} \right)\left( {{\left( 3x \right)}^{2}}-{{2}^{2}} \right)$
Again applying $\left( {{a}^{2}}-{{b}^{2}} \right)$ in the second bracket of the above expression then in the expression written in the second bracket, $a=3x$ and $b=2$ we get,
$\Rightarrow \left( {{\left( 3x \right)}^{2}}+{{2}^{2}} \right)\left( 3x+2 \right)\left( 3x-2 \right)$
Now, simplifying the expression written in the first bracket in the above expression we get,
$\Rightarrow \left( 9{{x}^{2}}+4 \right)\left( 3x+2 \right)\left( 3x-2 \right)$
Hence, we have factorized the given expression as $\left( 9{{x}^{2}}+4 \right)\left( 3x+2 \right)\left( 3x-2 \right)$.

Note:
To check if the factorization which we have done above is correct or not we are going to multiply all the factors which we have calculated above and then see either we are restoring the original expression or not.
The factors which we have found above for the given expression are:
$\left( 9{{x}^{2}}+4 \right)\left( 3x+2 \right)\left( 3x-2 \right)$
Multiplying the last two expressions written in the brackets we get,
$\begin{align}
  & \Rightarrow \left( 9{{x}^{2}}+4 \right)\left( {{\left( 3x \right)}^{2}}-6x+6x-{{2}^{2}} \right) \\
 & =\left( 9{{x}^{2}}+4 \right)\left( 9{{x}^{2}}-4 \right) \\
\end{align}$
Now, multiplying the above two expressions written inside the bracket we get,
$\begin{align}
  & \Rightarrow 9{{x}^{2}}\left( 9{{x}^{2}} \right)-4\left( 9{{x}^{2}} \right)+4\left( 9{{x}^{2}} \right)-4\left( 4 \right) \\
 & =81{{x}^{4}}-36{{x}^{2}}+36{{x}^{2}}-16 \\
 & =81{{x}^{4}}+0-16 \\
 & =81{{x}^{4}}-16 \\
\end{align}$
As you can see we got the same expression which is the original expression which is given in the problem so the factors which we have solved in the above solution are correct.