How do you factor ${x^{12}} + 8$?

Hint: Given an expression. We have to factorize the expression completely. First we will rewrite the expression as a sum of cubes. Then, we will apply the algebraic identity for the sum of cubes to the expression. Then, we will further factorize the expression by again applying the algebraic identities to each factor obtained. Then, write the result of the expression as a product of factors.

Complete step by step solution:
We are given the expression, ${x^{12}} + 8$. First we will try to get the form so that we can apply any algebraic identity to reduce the expression.

Now, we will rewrite the expression as a sum of cubes of two numbers.

$ \Rightarrow {x^{12}} + 8 = {\left( {{x^4}} \right)^3} + {2^3}$

Now, we will apply the algebraic identity ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - 2ab + {b^2}} \right)$to the expression.

By substituting the value of $a$ as ${x^4}$ and value of $b$ as $2$ in the expression ${a^3} + {b^3}$, we get:

$ \Rightarrow {\left( {{x^4}} \right)^3} + {2^3} = \left( {{x^4} + 2} \right)\left( {{{\left( {{x^4}} \right)}^2} - 2 \times {x^4} \times 2 + {2^2}} \right)$

On further simplifying the expression, we get:

$ \Rightarrow {\left( {{x^4}} \right)^3} + {2^3} = \left( {{x^4} + 2} \right)\left( {{x^8} - 4{x^4} + 4} \right)$

Now, we will apply the algebraic identity, ${a^4} + {b^4} = \left( {{a^2} - \sqrt 2 ab + {b^2}} \right)\left( {{a^2} + \sqrt 2 ab + {b^2}} \right)$

By substituting the value of $a$ as $x$ and value of $b$as $\sqrt[4]{2}$ in the expression ${a^4} + {b^4}$, we get:

$ \Rightarrow {x^4} + 2 = \left( {{x^2} - \sqrt 2 x\sqrt[4]{2} + {{\left( {\sqrt[4]{2}} \right)}^2}} \right)\left( {{x^2} + \sqrt 2 x\sqrt[4]{2} + {{\left( {\sqrt[4]{2}} \right)}^2}} \right)$

On further simplifying the expression, we get:

$ \Rightarrow {x^4} + 2 = \left( {{x^2} - \sqrt[4]{8}x + \sqrt 2 } \right)\left( {{x^2} + \sqrt[4]{8}x + \sqrt 2 } \right)$

Now, factorize the expression $\left( {{x^8} - 4{x^4} + 4} \right)$ using the algebraic identity,${a^4} - {a^2}{b^2} + {b^4} = \left( {{a^2} - \sqrt 3 ab + {b^2}} \right)\left( {{a^2} + \sqrt 3 ab + {b^2}} \right)$.

By substituting the value of $a$ as ${x^2}$ and value of $b$ as $\sqrt 2 $ into the expression, we get:

$ \Rightarrow \left( {{x^8} - 4{x^4} + 4} \right) = \left( {{x^2} - \sqrt 6 {x^2} + 2} \right)\left( {{x^2} - \sqrt 6 {x^2} + 2} \right)$

Therefore, factors of ${x^{12}} + 8$ are $\left( {{x^2} - \sqrt[4]{8}x + \sqrt 2 } \right)\left( {{x^2} + \sqrt[4]{8}x + \sqrt 2 } \right)\left( {{x^2} - \sqrt 6 {x^2} + 2} \right)\left( {{x^2} - \sqrt 6 {x^2} + 2} \right)$.

Note: The students please note that the factorization of a polynomial consists of writing a polynomial as a product of several polynomials. Students must remember to use algebraic identities to simplify and factorize the expression until the expression is not possible to reduce further. We have applied various algebraic identities such as

$ \Rightarrow {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - 2ab + {b^2}} \right)$
$ \Rightarrow {a^4} + {b^4} = \left( {{a^2} - \sqrt 2 ab + {b^2}} \right)\left( {{a^2} + \sqrt 2 ab + {b^2}} \right)$
$ \Rightarrow {a^4} - {a^2}{b^2} + {b^4} = \left( {{a^2} - \sqrt 3 ab + {b^2}} \right)\left( {{a^2} + \sqrt 3 ab + {b^2}} \right)$