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**Hint:**Assume \[\sqrt[3]{2+\sqrt{5}}=x\] and cube both the sides. Now, find the value of \[\left( 2-\sqrt{5} \right)\] in terms of x by rationalizing \[\left( 2+\sqrt{5} \right)\]. To rationalize \[2+\sqrt{5}\] multiply and divide it with \[\left( 2-\sqrt{5} \right)\]. Now, use the formula: - \[{{\left( x-\dfrac{1}{x} \right)}^{3}}={{x}^{3}}-\dfrac{1}{{{x}^{3}}}-3\left( x-\dfrac{1}{x} \right)\] and assume \[\left( x-\dfrac{1}{x} \right)=k\]. Form a cubic equation in k and solve this equation to get the value of k which will be our answer.

**Complete step by step solution:**

Here, we have been provided with the expression: - \[\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\] and we are asked to solve it. That means we have to find the numerical value of this expression.

Now, let us assume \[\sqrt[3]{2+\sqrt{5}}=x\], so on cubing both the sides we get,

\[\Rightarrow 2+\sqrt{5}={{x}^{3}}\] - (1)

\[\Rightarrow {{x}^{3}}=2+\sqrt{5}\]

Rationalizing the R.H.S. by multiplying \[\left( 2+\sqrt{5} \right)\] with \[\left( 2-\sqrt{5} \right)\] and to balance the relation dividing it with \[\left( 2-\sqrt{5} \right)\], we get,

\[\Rightarrow {{x}^{3}}=\left( 2+\sqrt{5} \right)\times \left( \dfrac{2-\sqrt{5}}{2-\sqrt{5}} \right)\]

Using the identity: - \[\left( 2+\sqrt{5} \right)\left( 2-\sqrt{5} \right)={{2}^{2}}-{{\left( \sqrt{5} \right)}^{2}}\], i.e., \[\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}\], we get,

\[\begin{align}

& \Rightarrow {{x}^{3}}=\dfrac{{{2}^{2}}-{{\left( \sqrt{5} \right)}^{2}}}{2-\sqrt{5}} \\

& \Rightarrow {{x}^{3}}=\dfrac{4-5}{2-\sqrt{5}} \\

& \Rightarrow {{x}^{3}}=\dfrac{-1}{2-\sqrt{5}} \\

\end{align}\]

By cross – multiplying we get,

\[\Rightarrow 2-\sqrt{5}=\dfrac{-1}{{{x}^{3}}}\] - (2)

Taking cube root both the sides, we get,

\[\Rightarrow \sqrt[3]{2-\sqrt{5}}=\dfrac{-1}{x}\]

Since, we have to find the value of \[\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\], so we have,

\[\Rightarrow \sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=x-\dfrac{1}{x}\]

That means we have to find the value of \[x-\dfrac{1}{x}\]. Let us find the whole cube of \[x-\dfrac{1}{x}\]. Using the formula: - \[{{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)\], we have,

\[\begin{align}

& \Rightarrow {{\left( x-\dfrac{1}{x} \right)}^{3}}={{x}^{3}}-\dfrac{1}{{{x}^{3}}}-3x\times \dfrac{1}{x}\left( x-\dfrac{1}{x} \right) \\

& \Rightarrow {{\left( x-\dfrac{1}{x} \right)}^{3}}={{x}^{3}}-\dfrac{1}{{{x}^{3}}}-3\left( x-\dfrac{1}{x} \right) \\

\end{align}\]

Assuming \[\left( x-\dfrac{1}{x} \right)=k\], we have,

\[\Rightarrow {{k}^{3}}={{x}^{3}}-\dfrac{1}{{{x}^{3}}}-3k\]

Substituting the value of \[{{x}^{3}}\] and \[\dfrac{-1}{{{x}^{3}}}\] in the above relation using equations (1) and (2), we get,

\[\begin{align}

& \Rightarrow {{k}^{3}}=\left( 2+\sqrt{5} \right)-\left( 2+\sqrt{5} \right)-3k \\

& \Rightarrow {{k}^{3}}=4-3k \\

& \Rightarrow {{k}^{3}}+3k-4=0 \\

\end{align}\]

Since, this is a cubic equation in k so we have to find one root by the hit – and – trial method. Now, considering k = 1, we have,

\[\Rightarrow {{k}^{3}}+3k-4={{\left( 1 \right)}^{3}}+3\left( 1 \right)-4=0\]

Therefore, we can say that k = 1 is a root of the equation \[{{k}^{3}}+3k-4=0\]. So, we can factorize this equation as: -

\[\begin{align}

& \Rightarrow {{k}^{3}}+3k-4=0 \\

& \Rightarrow {{k}^{2}}\left( k-1 \right)+k\left( k-1 \right)+4\left( k-1 \right)=0 \\

& \Rightarrow \left( k-1 \right)\left( {{k}^{2}}+k+4 \right)=0 \\

\end{align}\]

Now, we can see that \[{{k}^{2}}+k+4\] will not have a real solution when it will be equated with 0, so the only value of k will be 1. The reason that \[{{k}^{2}}+k+4\] will not have any real solution is that the discriminant value of this quadratic equation is less than 0.

\[\begin{align}

& \Rightarrow k=1 \\

& \Rightarrow x-\dfrac{1}{x}=1 \\

& \Rightarrow \sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1 \\

\end{align}\]

**Hence, the value of the given expression is 1.**

**Note:**One may note that the expression \[\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\] will have only a real solution and that is why only one value of k was considered. Note that without rationalization we will not be able to solve this question as it will not be possible for us to find the cube root of \[2+\sqrt{5}\] and \[2-\sqrt{5}\]. You must remember the conditions for a quadratic equation to have real and imaginary roots.

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