\sqrt{5} +2 and \sqrt{5}-2 are symmetric values that have special properties.

(\sqrt{5}+2 )(\sqrt{5}-2 ) = 1 and \dfrac{1}{ \sqrt{5}+2} = \sqrt{5}-2

which reminds us if it is possible to find a pair of numbers to which 3 power is equal to them.

2 is too small to be resulted by cube. In order to raise the integer term to a larger number, we can transform them to fractions with a denominator of which the cubic root is an integer. So we will start from the smallest denominator 8.

\sqrt{5}+2 = \dfrac{8}{8}(\sqrt{5}+2) =\dfrac{16+8\sqrt{5} }{8}

When the numerator 16+8\sqrt{5} can be represented by cube of a number, \sqrt{5}+2 will be cube of half of that number.

Let

16+8\sqrt{5} = (a+b\sqrt{5})^3

16-8\sqrt{5} = (a-b\sqrt{5})^3

in which a and b are positive

Addition of the two equations gives

(a+b\sqrt{5})^3+(a-b\sqrt{5})^3 = 32

Factoring the LHS using sum of cubes formula.

2a[(a+b\sqrt{5})^2+(a-b\sqrt{5})^2-a^2+5b^2)=32

a(2a^2+2b^2\cdot 5 - a^2+5b^2) = 16

Multiplying the two equations gives

(16+8\sqrt{5})(16-8\sqrt{5}) = (a+b\sqrt{5})^3 (a-b\sqrt{5})^3

Simplify the equation using difference of square formula

(a^2 - b^2\cdot 5)^3 = 16^2 - 8^2\cdot 5=-64

Taking cube root gives

A system of equations is obtained from (1) and (2)

\begin{cases} a^3 + 15ab^2 &=16 \\ a^2-5b^2 &=-4 \end{cases}

Solving the equation yields

a = 1, b=1

Therefore,

The cube root of

\sqrt[3]{16+8\sqrt{5}}=\sqrt[3]{(1+\sqrt{5})^3}=1+\sqrt{5}

\sqrt[3]{16-8\sqrt{5}}=\sqrt[3]{(1-\sqrt{5})^3}= \sqrt{5}-1

And

\sqrt{5} +2 = \Big( \dfrac{\sqrt{ 5}+1 }{2}\Big) ^3

\sqrt{5} - 2= \Big( \dfrac{\sqrt{ 5}-1 }{2}\Big) ^3

Substitute to the asking expression

(\sqrt[3]{\sqrt{5}+2 }+\sqrt[3]{\sqrt{5}-2 } )^{2014}

=\Big( \dfrac{\sqrt{ 5}+1 }{2}+ \dfrac{\sqrt{ 5}-1 }{2}\Big) ^{2014}

=\Big( \sqrt{5} \Big) ^{2014}

=5^{1007}