How do you factor 100 - 81t^6?

Jan 21, 2017

$\left(10 - 9 {t}^{3}\right) \left(10 + 9 {t}^{3}\right)$

Explanation:

You can recognize squares: 100 is 10^2, 81 is 9^2
Then $100 - 81 {t}^{6} = {10}^{2} - {\left(9 {t}^{3}\right)}^{2} = \left(10 - 9 {t}^{3}\right) \left(10 + 9 {t}^{3}\right)$

Jan 22, 2017

$100 - 81 {t}^{6}$

$= \left(10 - 9 {t}^{3}\right) \left(10 + 9 {t}^{3}\right)$

$= \left(\sqrt[3]{10} - \sqrt[3]{9} t\right) \left(\sqrt[3]{100} + \sqrt[3]{90} t + \sqrt[3]{81} {t}^{2}\right) \left(\sqrt[3]{10} + \sqrt[3]{9} t\right) \left(\sqrt[3]{100} - \sqrt[3]{90} t + \sqrt[3]{81} {t}^{2}\right)$

Explanation:

The difference of squares identity can be written:

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

The difference of cubes identity can be written:

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

The sum of cubes identity can be written:

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

Note also that:

$\sqrt[3]{a} \sqrt[3]{b} = \sqrt[3]{a b}$

Hence we find:

$100 - 81 {t}^{6}$

$= {10}^{2} - {\left(9 {t}^{3}\right)}^{2}$

$= \left(10 - 9 {t}^{3}\right) \left(10 + 9 {t}^{3}\right)$

$= \left({\left(\sqrt[3]{10}\right)}^{3} - {\left(\sqrt[3]{9} t\right)}^{3}\right) \left({\left(\sqrt[3]{10}\right)}^{3} + {\left(\sqrt[3]{9} t\right)}^{3}\right)$

$= \left(\sqrt[3]{10} - \sqrt[3]{9} t\right) \left(\sqrt[3]{100} + \sqrt[3]{90} t + \sqrt[3]{81} {t}^{2}\right) \left(\sqrt[3]{10} + \sqrt[3]{9} t\right) \left(\sqrt[3]{100} - \sqrt[3]{90} t + \sqrt[3]{81} {t}^{2}\right)$