How do you factor the expression ${{a}^{4}}-81$ ?

Hint: We can use the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to this question .We can write ${{a}^{4}}$ as ${{\left( {{a}^{2}} \right)}^{2}}$ and 81 as ${{9}^{2}}$ then apply the formula to factorize the equation.

Complete step by step answer:
We know that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
We can write ${{a}^{4}}-81$ as ${{\left( {{a}^{2}} \right)}^{2}}-{{9}^{2}}$ and apply the above formula. That gives us
$\Rightarrow {{\left( {{a}^{2}} \right)}^{2}}-{{9}^{2}}=\left( {{a}^{2}}+9 \right)\left( {{a}^{2}}-9 \right)$ …eq1
Now we have $\left( {{a}^{2}}+9 \right)\left( {{a}^{2}}-9 \right)$ we can write $\left( {{a}^{2}}-9 \right)$ as ${{a}^{2}}-{{3}^{2}}$ and apply the formula
$\left( {{a}^{2}}-9 \right)=\left( a+3 \right)\left( a-3 \right)$
Replacing $\left( {{a}^{2}}-9 \right)$ with $\left( a+3 \right)\left( a-3 \right)$ in eq1 we get
${{a}^{4}}-81=\left( {{a}^{2}}+9 \right)\left( a+3 \right)\left( a-3 \right)$
This is one way we can factorize the equation.

Note: Another way of solving the problem is using binomial theorem. We know that formula
For ${{a}^{n}}-{{b}^{n}}=\left( a-b \right)\left( {{a}^{n-1}}+{{a}^{n-2}}b+{{a}^{n-3}}{{b}^{2}}+.......+{{b}^{n-2}}a+{{b}^{n-1}} \right)$ where n is a positive integer we can apply this formula to solve this question we have ${{a}^{4}}-81$ we can replace 81 as ${{3}^{4}}$ in the equation the problem will be ${{a}^{4}}-{{3}^{4}}$ . Now if we replace b with 3 and n with 4 in the binomial theorem
We will get
$\Rightarrow \left( {{a}^{4}}-{{3}^{4}} \right)=\left( a-3 \right)\left( {{a}^{3}}+3{{a}^{2}}+9a+27 \right)$ …eq2
Now we have to factorize the second term in the equation that is $\left( {{a}^{3}}+3{{a}^{2}}+9a+27 \right)$
For that we can take common ${{a}^{2}}$ in the first half of the term and take 9 common in second half of the term
$\Rightarrow \left( {{a}^{3}}+3{{a}^{2}}+9a+27 \right)={{a}^{2}}\left( a+3 \right)+9\left( a+3 \right)$
Taking $a+3$ in the equation
$\Rightarrow {{a}^{2}}\left( a+3 \right)+9\left( a+3 \right)=\left( {{a}^{2}}+9 \right)\left( a+3 \right)$

By replacing $\left( {{a}^{3}}+3{{a}^{2}}+9a+27 \right)$ with $\left( {{a}^{2}}+9 \right)\left( a+3 \right)$ in eq2 we get
$\Rightarrow \left( {{a}^{4}}-{{3}^{4}} \right)=\left( a-3 \right)\left( a+3 \right)\left( {{a}^{2}}+9 \right)$
Remember the above binomial formula is valid when n is a positive integer. The first method is the shortest. This method will come to use while solving ${{a}^{n}}-{{b}^{n}}$ where n is equal to ${{2}^{m}};m\ge 0$
For example we can write
(1) ${{a}^{8}}-{{b}^{8}}=\left( a-b \right)\left( a+b \right)\left( {{a}^{2}}-{{b}^{2}} \right)\left( {{a}^{2}}+{{b}^{2}} \right)\left( {{a}^{4}}+{{b}^{4}} \right)$
(2) ${{a}^{4}}-{{b}^{4}}=\left( a-b \right)\left( a+b \right)\left( {{a}^{2}}+{{b}^{2}} \right)$
Whenever n is equal to ${{2}^{m}};m\ge 0$ we can use the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$
 m times to factorize the equation.