How do you factor ${x^6} + 125$?

Hint:To order to determine the factors of the above equation ,first rewrite the equation by writing ${x^6}$ as ${\left( {{x^3}} \right)^2}$ using property of exponent \[{a^{m \times n}} = {\left( {{a^m}} \right)^n}\] and $125$ as $5 \times 5 \times 5 = {5^3}$ and then use the formula of sum of two cubes $\left( {{A^3} + {B^3}} \right) = \left( {A + B} \right)\left( {{A^2} - A.B + {B^2}} \right)$ by considering $A$ as ${x^2}$ and $B$ as ${5^3}$.To factorise a quartic part Solve it as $\left( {{a^2} - kab + {b^2}} \right)\left( {{a^2} + kab + {b^2}} \right) = {a^4} + \left( {2 - {k^2}} \right){a^2}{b^2} + {b^4}$by comparing .You’ll get your required factors.

Complete step by step solution:
$\left( {{A^3} - {B^3}} \right) = \left( {A - B} \right)\left( {{A^2} + A.B + {B^2}} \right)$
We are given a mathematical function having variable ($x$),let it be $f(x)$
$f(x) = {x^6} + 125$
To split this expression into its factors ,we will be writing $125$as$5 \times 5 \times 5 = {5^3}$ and using the property of exponents that \[{a^{m \times n}} = {\left( {{a^m}} \right)^n}\]So writing
${x^6}$ as ${\left( {{x^3}} \right)^2}$.$f(x)$ will now become,
$f(x) = {\left( {{x^2}} \right)^3} + {5^3}$
Now we’ll be using the formula of sum of two cube number
$\left( {{A^3} + {B^3}} \right) = \left( {A + B} \right)\left( {{A^2} - A.B + {B^2}} \right)$by considering $A$ as ${x^2}$ and $B$ as ${5^3}$
$
f(x) = \left( {{x^2} + 5} \right)\left( {{{({x^2})}^2} - 5({x^2}) + {5^2}} \right) \\
= \left( {{x^2} + 5} \right)\left( {{x^4} - 5{x^2} + 25} \right) \\
$----(1)
To factorise the Quartic part of the expression we’ll use a different aspect as
$\left( {{a^2} - kab + {b^2}} \right)\left( {{a^2} + kab + {b^2}} \right) = {a^4} + \left( {2 - {k^2}}
\right){a^2}{b^2} + {b^4}$
So, now comparing the result mentioned above ${a^4} + \left( {2 - {k^2}} \right){a^2}{b^2} +
{b^4}$ with our original expression
$
a = x \\
b = \sqrt 5 \\
$
We find that:
$\left( {{x^2} - k\sqrt 5 x + 5} \right)\left( {{x^2} + k\sqrt 5 x + 5} \right) = {x^4} + \left( {2 - {k^2}} \right)5{x^2} + 25$-----(2)
Equating coefficients of ${x^2}$ of the above with the same in equation (1),we get,
$
\left( {2 - {k^2}} \right)5 = - 5 \\
\left( {2 - {k^2}} \right) = - 1 \\
{k^2} = 3 \\
k = \pm \sqrt 3 \\
$
So, putting $k = \pm \sqrt 3 $ in the equation (2)
\[
{x^4} - 5{x^2} + 25 = \left( {{x^2} - \sqrt 3 .\sqrt 5 x + 5} \right)\left( {{x^2} + \sqrt 3 .\sqrt 5 x + 5} \right) \\
= \left( {{x^2} - \sqrt {15} x + 5} \right)\left( {{x^2} + \sqrt {15} x + 5} \right) \\
\]
Putting back the value of \[{x^4} - 5{x^2} + 25\]in equation (1),we finally get,
\[f(x) = \left( {{x^2} + 5} \right)\left( {{x^2} - \sqrt {15} x + 5} \right)\left( {{x^2} + \sqrt {15} x + 5} \right)\]
Therefore, we have successfully factorized $f(x)$ as\[\left( {{x^2} + 5} \right)\left( {{x^2} - \sqrt {15} x + 5} \right)\left( {{x^2} + \sqrt {15} x + 5} \right)\].

Additional Information:
Cubic Equation: A cubic equation is a equation which can be represented in the form of $a{x^3} + b{x^2}cx + d$ where $x$ is the unknown variable and a,b,c,d are the numbers known where $a \ne 0$.If $a = 0$ then the equation will become a quadratic equation and will no longer be cubic

Note:
1. One must be careful while calculating the answer as calculation error may come.
2.Don’t forget to compare the given cubic equation with the standard one every time.