# What is the value of (2+root5)^1/3 + (2-root5)^1/3 ?

Feb 3, 2018

The value is $- 2$

#### Explanation:

Let $x = {\left(2 + \sqrt{5}\right)}^{\frac{1}{3}} + {\left(2 - \sqrt{5}\right)}^{\frac{1}{3}}$ then

${x}^{3} = {\left\{{\left(2 + \sqrt{5}\right)}^{\frac{1}{3}} + {\left(2 - \sqrt{5}\right)}^{\frac{1}{3}}\right\}}^{3}$

Reminder:

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

and let $a = \left(2 + \sqrt{5}\right) , b = \left(2 - \sqrt{5}\right) \therefore a b = 4 - 5 = - 1$

$\therefore {x}^{3} = {\left(2 + \sqrt{5}\right)}^{3 \cdot \frac{1}{3}} + {\left(2 - \sqrt{5}\right)}^{3 \cdot \frac{1}{3}} + 3 \left(2 + \sqrt{5}\right) \left(2 - \sqrt{5}\right) \left(2 + \sqrt{5} + 2 - \sqrt{5}\right)$

or ${x}^{3} = 2 + \cancel{\sqrt{5}} + 2 - \cancel{\sqrt{5}} + 3 \left(4 - 5\right) \left(2 + \cancel{\sqrt{5}} + 2 - \cancel{\sqrt{5}}\right)$

or ${x}^{3} = 4 + 3 \left(- 1\right) \left(4\right) \mathmr{and} {x}^{3} = 4 - 12 \mathmr{and} {x}^{3} = - 8$ or

$x = {\left(- 8\right)}^{\frac{1}{3}} = - 2$ . The value is $- 2$ [Ans]