Question

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

Answer 1

1

Answer 2

Let $x = \left(2+\sqrt{5}\right)^{\frac{1}{3}}+\left(2-\sqrt{5}\right)^{\frac{1}{3}}$. Cube both sides using the formula $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$. Let $a = \left(2+\sqrt{5}\right)^{\frac{1}{3}}$ and $b = \left(2-\sqrt{5}\right)^{\frac{1}{3}}$. Then $x^3 = a^3 + b^3 + 3ab(a+b)$.

$a^3 = \left(\left(2+\sqrt{5}\right)^{\frac{1}{3}}\right)^3 = 2+\sqrt{5}$ $b^3 = \left(\left(2-\sqrt{5}\right)^{\frac{1}{3}}\right)^3 = 2-\sqrt{5}$ $ab = \left(2+\sqrt{5}\right)^{\frac{1}{3}} \left(2-\sqrt{5}\right)^{\frac{1}{3}} = \left((2+\sqrt{5})(2-\sqrt{5})\right)^{\frac{1}{3}} = (4 - 5)^{\frac{1}{3}} = (-1)^{\frac{1}{3}} = -1$

Substituting these values into the cubed equation: $x^3 = (2+\sqrt{5}) + (2-\sqrt{5}) + 3(-1)(x)$ $x^3 = 4 - 3x$ Rearrange the equation to form a cubic equation: $x^3 + 3x - 4 = 0$

We need to find the real root(s) of this equation. We can test integer divisors of the constant term (-4), which are $\pm 1, \pm 2, \pm 4$. Let's test $x=1$: $1^3 + 3(1) - 4 = 1 + 3 - 4 = 0$. So, $x=1$ is a root.

We can factor the cubic polynomial using the root $x=1$. $(x-1)$ is a factor. Using polynomial division or synthetic division: $(x^3 + 3x - 4) \div (x-1) = x^2 + x + 4$. So, the equation is $(x-1)(x^2 + x + 4) = 0$. The roots are $x=1$ or $x^2 + x + 4 = 0$. Consider the quadratic equation $x^2 + x + 4 = 0$. The discriminant is $\Delta = b^2 - 4ac = 1^2 - 4(1)(4) = 1 - 16 = -15$. Since the discriminant is negative, the quadratic equation has two complex conjugate roots. The roots are $x = \frac{-1 \pm \sqrt{-15}}{2} = \frac{-1 \pm i\sqrt{15}}{2}$. The only real root of $x^3 + 3x - 4 = 0$ is $x=1$.

Since the expression $\left(2+\sqrt{5}\right)^{\frac{1}{3}}+\left(2-\sqrt{5}\right)^{\frac{1}{3}}$ must be a real number, its value is the real root of the cubic equation, which is 1.