How do you factor x^3 + 4 = 0?

Feb 13, 2016

${x}^{3} + 4 = \left(x + {4}^{\frac{1}{3}}\right) \left({x}^{2} - {4}^{\frac{1}{3}} x + {4}^{\frac{2}{3}}\right)$

$= \left(x + {4}^{\frac{1}{3}}\right) \left(x - {2}^{- \frac{1}{3}} \left(1 + \sqrt{3} i\right)\right) \left(x - {2}^{- \frac{1}{3}} \left(1 - \sqrt{3} i\right)\right)$

Explanation:

Using the sum of cubes formula ${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$ we have:

${x}^{3} + 4 = {x}^{3} + {\left({4}^{\frac{1}{3}}\right)}^{3}$

$= \left(x + {4}^{\frac{1}{3}}\right) \left({x}^{2} - {4}^{\frac{1}{3}} x + {4}^{\frac{2}{3}}\right)$

If we wish to restrict ourselves to real numbers, we are done. If we allow for complex numbers, then we may continue using the quadratic formula to factor ${x}^{2} - {4}^{\frac{1}{3}} x + {4}^{\frac{2}{3}}$

$x = \frac{{4}^{\frac{1}{3}} \pm \sqrt{{4}^{\frac{2}{3}} - 4 \left(1\right) \left({4}^{\frac{2}{3}}\right)}}{2}$

$= {2}^{- \frac{1}{3}} \left(1 \pm \sqrt{- 3}\right)$

$= {2}^{- \frac{1}{3}} \left(1 \pm \sqrt{3} i\right)$

Thus

${x}^{3} + 4 = \left(x + {4}^{\frac{1}{3}}\right) \left(x - {2}^{- \frac{1}{3}} \left(1 + \sqrt{3} i\right)\right) \left(x - {2}^{- \frac{1}{3}} \left(1 - \sqrt{3} i\right)\right)$