How do you factor -49 + y^6?

May 31, 2015

${y}^{6} - 49 = \left({y}^{3} - 7\right) \left({y}^{3} + 7\right)$

May 31, 2015

${y}^{6} - 49 = \left({y}^{3} + 7\right) \left({y}^{3} - 7\right)$

Problem: Factor $- 49 + {y}^{6}$.

Rewrite the equation as ${y}^{6} - 49$.

This is an example of the difference of squares:

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

$a = {y}^{3}$
$b = 7$

${\left({y}^{3}\right)}^{2} - {7}^{2} = \left({y}^{3} + 7\right) \left({y}^{3} - 7\right)$

May 31, 2015

$- 49 + {y}^{6} = {y}^{6} - 49 = {\left({y}^{3}\right)}^{2} - {7}^{2} = \left({y}^{3} - 7\right) \left({y}^{3} + 7\right)$

This is as far as you can go with rational coefficients.

If you are allowed real coefficients:

${y}^{3} - 7 = {y}^{3} - {\left(\sqrt[3]{7}\right)}^{3}$

$= \left(y - \sqrt[3]{7}\right) \left({y}^{2} + \left(\sqrt[3]{7}\right) y + {\left(\sqrt[3]{7}\right)}^{2}\right)$

and

${y}^{3} + 7 = {y}^{3} + {\left(\sqrt[3]{7}\right)}^{3}$

$= \left(y + \sqrt[3]{7}\right) \left({y}^{2} - \left(\sqrt[3]{7}\right) y + {\left(\sqrt[3]{7}\right)}^{2}\right)$

For shorthand I will write $\rho = \sqrt[3]{7}$

$- 49 + {y}^{6}$

$= \left(y - \rho\right) \left({y}^{2} + \rho y + {\rho}^{2}\right) \left(y + \rho\right) \left({y}^{2} - \rho y + {\rho}^{2}\right)$