# How do you factor x^6+10x^3+25?

##### 1 Answer
Sep 2, 2016

${x}^{6} + 10 {x}^{3} + 25 = {\left({x}^{3} + 5\right)}^{2} = {\left(x + \sqrt[3]{5}\right)}^{2} {\left({x}^{2} - \left(\sqrt[3]{5}\right) x + \sqrt[3]{25}\right)}^{2}$

#### Explanation:

The sum of cubes identity can be written:

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

Use this with $a = x$ and $b = \sqrt[3]{5}$ as follows:

${x}^{6} + 10 {x}^{3} + 25 = {\left({x}^{3}\right)}^{2} + 2 \left(5\right) \left({x}^{3}\right) + {\left(5\right)}^{2}$

$\textcolor{w h i t e}{{x}^{6} + 10 {x}^{3} + 25} = {\left({x}^{3} + 5\right)}^{2}$

$\textcolor{w h i t e}{{x}^{6} + 10 {x}^{3} + 25} = {\left({x}^{3} + {\left(\sqrt[3]{5}\right)}^{3}\right)}^{2}$

$\textcolor{w h i t e}{{x}^{6} + 10 {x}^{3} + 25} = {\left(\left(x + \sqrt[3]{5}\right) \left({x}^{2} - x \left(\sqrt[3]{5}\right) + {\left(\sqrt[3]{5}\right)}^{2}\right)\right)}^{2}$

$\textcolor{w h i t e}{{x}^{6} + 10 {x}^{3} + 25} = {\left(x + \sqrt[3]{5}\right)}^{2} {\left({x}^{2} - \left(\sqrt[3]{5}\right) x + \sqrt[3]{25}\right)}^{2}$