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# How to solve $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$?

Last updated date: 16th Jul 2024
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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.