How do you factorize ${x^4} - 8{x^2} + 16$?

Hint: A difference in two perfect squares defines that there should be two terms, where the sign between the two terms as a minus sign and both the two terms contain perfect squares. The result after the factorization of the difference in two perfect squares should contain two binomial terms. One binomial term contains the sum of two terms whereas the other contains the difference of two terms.

$\left( {{a^2} - {b^2}} \right) = \left( {a + b} \right)\left( {a - b} \right)$

Complete step by step answer:
To solve the given question in a simple way, we will choose a variable like $u$
Therefore,
Let $u = {x^2}$
Now, you have ${x^4} - 8{x^2} + 16$
This is equivalent to ${\left( {{x^2}} \right)^2} - 8{x^2} + 16$
You can now insert the other variable i.e., $u$ in place of ${x^2}$
You get ${u^2} - 8u + 16$
Now, you can factor this like a normal polynomial.
$\left( {u - 4} \right)\left( {u - 4} \right)$
Or
${\left( {u - 4} \right)^2}$
Finally, you plug out $u$ back from the expression
The factored form will be: ${\left( {{x^2} - 4} \right)^2}$

Additional Information:
The given question can also be solved in the following manner:
$
   = \left( {{x^4} - 8{x^2} + 16} \right) \\
   = \left( {{x^2} - 4} \right)\left( {{x^2} - 4} \right) \\
   = \left( {x + 2} \right)\left( {x - 2} \right)\left( {x + 2} \right)\left( {x - 2} \right) \\
   = {\left( {x + 2} \right)^2}{\left( {x - 2} \right)^2} \\
   = {\left( {{x^2} - 4} \right)^2} \\
 $

Note: In mathematics and computer algebra, factorization of polynomial or polynomial factorization expresses a polynomial with coefficient in a given field or in the integers as the product of irreducible factors with coefficients in the same domain. Polynomial factorization is one of the fundamental components of computer algebra systems.