Answer
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Hint: To solve the question given above, we will first assume that the value of $\left( \sqrt{288}-\sqrt{119} \right)$ is x and the value of $\left( \sqrt{288}+\sqrt{119} \right)$ be y. Then we will calculate the value of ${{x}^{3}}$ and ${{y}^{3}}$and then we will take the square root of the product to obtain the answer.
Complete step-by-step solution
Before solving the question given above, we will assume that the expression ${{\left( \sqrt{288}+\sqrt{119} \right)}^{\dfrac{3}{2}}}{{\left( \sqrt{288}+\sqrt{119} \right)}^{\dfrac{3}{2}}}$ has value equal to P. Thus, we have the following equation:
$P={{\left( \sqrt{288}-\sqrt{119} \right)}^{\dfrac{3}{2}}}{{\left( \sqrt{288}+\sqrt{119} \right)}^{\dfrac{3}{2}}}.....\left( 1 \right)$
Now, we will assume that the value of $\left( \sqrt{288}-\sqrt{119} \right)$ be x. Thus we have the following equation:
$x=\sqrt{288}-\sqrt{119}......\left( 2 \right)$
Now, we will cube both sides of the equation. After doing this we will get following equation:
${{x}^{3}}={{\left( \sqrt{288}-\sqrt{119} \right)}^{3}}$
In the above equation, we are going to use an identity which is shown below:
${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$
Thus, we will get:
${{x}^{3}}={{\left( \sqrt{288} \right)}^{3}}-{{\left( \sqrt{119} \right)}^{3}}-3{{\left( \sqrt{288} \right)}^{2}}\left( \sqrt{119} \right)+3\left( \sqrt{288} \right){{\left( \sqrt{119} \right)}^{2}}$
$\Rightarrow {{x}^{3}}={{\left( \sqrt{288} \right)}^{3}}-{{\left( \sqrt{119} \right)}^{3}}-\left[ 3\left( 288 \right)\left( \sqrt{119} \right) \right]+\left[ 3\left( \sqrt{288} \right)\left( 119 \right) \right]$
Now, we will rearrange the above equation and we will get:
\[\begin{align}
& \Rightarrow {{x}^{3}}={{\left( \sqrt{288} \right)}^{3}}+\left[ 3\left( \sqrt{288} \right)\left( 119 \right) \right]-{{\left( \sqrt{119} \right)}^{3}}-\left[ 3\left( 288 \right)\left( \sqrt{119} \right) \right] \\
& \Rightarrow {{x}^{3}}=\sqrt{288}\left( {{\left( \sqrt{288} \right)}^{2}}+\left[ 3\left( 119 \right) \right] \right)-\sqrt{119}\left( {{\left( \sqrt{119} \right)}^{2}}+\left[ 3\left( 288 \right) \right] \right) \\
\end{align}\]
Now, we will square the terms which has the power 2 and therefore we will get:
\[\begin{align}
& \Rightarrow {{x}^{3}}=\sqrt{288}\left( 288+\left[ 3\left( 119 \right) \right] \right)-\sqrt{119}\left( 119+\left[ 3\left( 288 \right) \right] \right) \\
& \Rightarrow {{x}^{3}}=\sqrt{288}\left( 288+\left[ 3\left( 119 \right) \right] \right)-\sqrt{119}\left( 119+\left[ 3\left( 288 \right) \right] \right) \\
& \Rightarrow {{x}^{3}}=\sqrt{288}\left( 645 \right)-\sqrt{119}\left( 983 \right)\text{ }.........\left( 3 \right) \\
\end{align}\]
Now, let us assume that the value of $\left( \sqrt{288}+\sqrt{119} \right)$ be y. Thus we have:
$y=\sqrt{288}+\sqrt{119}$
On cubing both sides we get:
${{y}^{3}}={{\left( \sqrt{288}+\sqrt{119} \right)}^{3}}$
Now, we will use another identity in above equation which is shown below:
${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}$
Thus, we will get:
${{y}^{3}}={{\left( \sqrt{288} \right)}^{3}}+{{\left( \sqrt{119} \right)}^{3}}+3{{\left( \sqrt{288} \right)}^{2}}\left( \sqrt{119} \right)+3\left( \sqrt{288} \right){{\left( \sqrt{119} \right)}^{2}}$
$\Rightarrow {{y}^{3}}={{\left( \sqrt{288} \right)}^{3}}+{{\left( \sqrt{119} \right)}^{3}}+\left[ 3\left( 288 \right)\left( \sqrt{119} \right) \right]+\left[ 3\left( \sqrt{288} \right)\left( 119 \right) \right]$
Now, we will rearrange the above equation and we will get:
\[\begin{align}
& \Rightarrow {{y}^{3}}={{\left( \sqrt{288} \right)}^{3}}+\left[ 3\left( \sqrt{288} \right)\left( 119 \right) \right]+{{\left( \sqrt{119} \right)}^{3}}+\left[ 3\left( 288 \right)\left( \sqrt{119} \right) \right] \\
& \Rightarrow {{y}^{3}}=\sqrt{288}\left( {{\left( \sqrt{288} \right)}^{2}}+\left[ 3\left( 119 \right) \right] \right)+\sqrt{119}\left( {{\left( \sqrt{119} \right)}^{2}}+\left[ 3\left( 288 \right) \right] \right) \\
\end{align}\]
Now, we will square the terms which has the power 2 and therefore we will get:
\[\begin{align}
& \Rightarrow {{y}^{3}}=\sqrt{288}\left( 288+\left[ 3\left( 119 \right) \right] \right)+\sqrt{119}\left( 119+\left[ 3\left( 288 \right) \right] \right) \\
& \Rightarrow {{y}^{3}}=\sqrt{288}\left( 288+\left[ 3\left( 119 \right) \right] \right)+\sqrt{119}\left( 119+\left[ 3\left( 288 \right) \right] \right) \\
& \Rightarrow {{y}^{3}}=\sqrt{288}\left( 645 \right)+\sqrt{119}\left( 983 \right)\text{ }.........\left( 4 \right) \\
\end{align}\]
Now, we will multiply the equation $\left( 3 \right)$ and $\left( 4 \right)$. After doing this, we will get:
${{x}^{3}}{{y}^{3}}=\left( \sqrt{288}\left( 645 \right)-\sqrt{119}\left( 983 \right) \right)\left( \sqrt{288}\left( 645 \right)+\sqrt{119}\left( 983 \right) \right)$
Here, we will use the identity on the RHS: $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
$\begin{align}
& \Rightarrow {{x}^{3}}{{y}^{3}}={{\left( \sqrt{288}\left( 645 \right) \right)}^{2}}-{{\left( \sqrt{119}\left( 983 \right) \right)}^{2}} \\
& \Rightarrow {{x}^{3}}{{y}^{3}}=288{{\left( 645 \right)}^{2}}-119{{\left( 983 \right)}^{2}} \\
\end{align}$
Now. we will expand 645 and 983
$\Rightarrow {{x}^{3}}{{y}^{3}}=288{{\left( 288+\left( 3\times 119 \right) \right)}^{2}}-119{{\left( 119+\left( 3\times 288 \right) \right)}^{2}}$
Now, we will apply the identity that is ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ :
\[\Rightarrow {{x}^{3}}{{y}^{3}}=288\left( {{\left( 288 \right)}^{2}}+{{\left( 3\times 119 \right)}^{2}}+2\left( 288 \right)\left( 3\times 119 \right) \right)-119\left( {{\left( 119 \right)}^{2}}+{{\left( 3\times 288 \right)}^{2}}+2\left( 3\times 288 \right)\left( 119 \right) \right)\]
Now, we will multiply 288 and 119 into the brackets:
\[\begin{align}
& \Rightarrow {{x}^{3}}{{y}^{3}}=\left( {{\left( 288 \right)}^{3}}+288{{\left( 3\times 119 \right)}^{2}}+2{{\left( 288 \right)}^{2}}\left( 3\times 119 \right) \right)-\left( {{\left( 119 \right)}^{3}}+119{{\left( 3\times 288 \right)}^{2}}+2\left( 3\times 288 \right){{\left( 119 \right)}^{2}} \right) \\
& \Rightarrow {{x}^{3}}{{y}^{3}}=\left( {{\left( 288 \right)}^{3}}+9\times 288\times {{\left( 119 \right)}^{2}}+6{{\left( 288 \right)}^{2}}119 \right)-\left( {{\left( 119 \right)}^{3}}+119\times 9\times {{\left( 288 \right)}^{2}}+6\times 288{{\left( 119 \right)}^{2}} \right) \\
\end{align}\]
Now, we will arrange the above terms:
\[\begin{align}
& \Rightarrow {{x}^{3}}{{y}^{3}}={{\left( 288 \right)}^{3}}-{{\left( 119 \right)}^{3}}+9\times 288\times {{\left( 119 \right)}^{2}}-6\times 288{{\left( 119 \right)}^{2}}+6{{\left( 288 \right)}^{2}}119-119\times 9\times {{\left( 288 \right)}^{2}} \\
& \Rightarrow {{x}^{3}}{{y}^{3}}={{\left( 288 \right)}^{3}}-{{\left( 119 \right)}^{3}}+288{{\left( 119 \right)}^{2}}\left( 9-6 \right)+{{\left( 288 \right)}^{2}}119\left( 6-9 \right) \\
& \Rightarrow {{x}^{3}}{{y}^{3}}={{\left( 288 \right)}^{3}}-{{\left( 119 \right)}^{3}}+3{{\left( 119 \right)}^{2}}288-3{{\left( 288 \right)}^{2}}119 \\
\end{align}\]
Now, we know that: ${{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}={{\left( a-b \right)}^{3}}$
Therefore: $\Rightarrow {{x}^{3}}{{y}^{3}}={{\left( 288-119 \right)}^{3}}\Rightarrow {{x}^{3}}{{y}^{3}}={{\left( 169 \right)}^{3}}\text{ }.......\text{ Equation 5}$
Now, we know that
$P={{\left( \sqrt{288}-\sqrt{119} \right)}^{\dfrac{3}{2}}}{{\left( \sqrt{288}+\sqrt{119} \right)}^{\dfrac{3}{2}}}$
$\Rightarrow P={{x}^{\dfrac{3}{2}}}{{y}^{\dfrac{3}{2}}}$
$\Rightarrow P={{\left( {{x}^{3}} \right)}^{\dfrac{1}{2}}}{{\left( {{y}^{3}} \right)}^{\dfrac{1}{2}}}$
Now, we will use another identity: ${{a}^{x}}\times {{b}^{x}}={{\left( ab \right)}^{x}}$
Thus, we will get:
$\Rightarrow P={{\left( {{x}^{3}}{{y}^{3}} \right)}^{\dfrac{1}{2}}}$
$\Rightarrow P={{\left( {{169}^{3}} \right)}^{\dfrac{1}{2}}}$ (from equation 5)
$\Rightarrow P={{\left[ {{\left( 13\times 13 \right)}^{3}} \right]}^{\dfrac{1}{2}}}$
$\Rightarrow P={{\left( 13\times 13 \right)}^{\dfrac{3}{2}}}$
$\Rightarrow P={{\left( 13 \right)}^{3}}=13\times 13\times 13$
$\Rightarrow P=2197$
Thus, value of ${{\left( \sqrt{288}-\sqrt{119} \right)}^{\dfrac{3}{2}}}{{\left( \sqrt{288}+\sqrt{119} \right)}^{\dfrac{3}{2}}}=2197$
Note: The above question can also be solved alternatively as shown below:
$P={{\left( \sqrt{288}-\sqrt{119} \right)}^{\dfrac{3}{2}}}{{\left( \sqrt{288}+\sqrt{119} \right)}^{\dfrac{3}{2}}}$
Here, we will use identity: ${{a}^{x}}\times {{b}^{x}}={{\left( ab \right)}^{x}}$
$\Rightarrow P={{\left[ \left( \sqrt{288}-\sqrt{119} \right)\left( \sqrt{288}+\sqrt{119} \right) \right]}^{\dfrac{3}{2}}}$
Now, we will use identity: $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
$\Rightarrow P={{\left( 288-119 \right)}^{\dfrac{3}{2}}}$
$\Rightarrow P={{\left( 169 \right)}^{\dfrac{3}{2}}}$
$\Rightarrow P={{\left( 13\times 13 \right)}^{\dfrac{3}{2}}}$
$\Rightarrow P={{13}^{3}}$
$\Rightarrow P=2197$
Complete step-by-step solution
Before solving the question given above, we will assume that the expression ${{\left( \sqrt{288}+\sqrt{119} \right)}^{\dfrac{3}{2}}}{{\left( \sqrt{288}+\sqrt{119} \right)}^{\dfrac{3}{2}}}$ has value equal to P. Thus, we have the following equation:
$P={{\left( \sqrt{288}-\sqrt{119} \right)}^{\dfrac{3}{2}}}{{\left( \sqrt{288}+\sqrt{119} \right)}^{\dfrac{3}{2}}}.....\left( 1 \right)$
Now, we will assume that the value of $\left( \sqrt{288}-\sqrt{119} \right)$ be x. Thus we have the following equation:
$x=\sqrt{288}-\sqrt{119}......\left( 2 \right)$
Now, we will cube both sides of the equation. After doing this we will get following equation:
${{x}^{3}}={{\left( \sqrt{288}-\sqrt{119} \right)}^{3}}$
In the above equation, we are going to use an identity which is shown below:
${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$
Thus, we will get:
${{x}^{3}}={{\left( \sqrt{288} \right)}^{3}}-{{\left( \sqrt{119} \right)}^{3}}-3{{\left( \sqrt{288} \right)}^{2}}\left( \sqrt{119} \right)+3\left( \sqrt{288} \right){{\left( \sqrt{119} \right)}^{2}}$
$\Rightarrow {{x}^{3}}={{\left( \sqrt{288} \right)}^{3}}-{{\left( \sqrt{119} \right)}^{3}}-\left[ 3\left( 288 \right)\left( \sqrt{119} \right) \right]+\left[ 3\left( \sqrt{288} \right)\left( 119 \right) \right]$
Now, we will rearrange the above equation and we will get:
\[\begin{align}
& \Rightarrow {{x}^{3}}={{\left( \sqrt{288} \right)}^{3}}+\left[ 3\left( \sqrt{288} \right)\left( 119 \right) \right]-{{\left( \sqrt{119} \right)}^{3}}-\left[ 3\left( 288 \right)\left( \sqrt{119} \right) \right] \\
& \Rightarrow {{x}^{3}}=\sqrt{288}\left( {{\left( \sqrt{288} \right)}^{2}}+\left[ 3\left( 119 \right) \right] \right)-\sqrt{119}\left( {{\left( \sqrt{119} \right)}^{2}}+\left[ 3\left( 288 \right) \right] \right) \\
\end{align}\]
Now, we will square the terms which has the power 2 and therefore we will get:
\[\begin{align}
& \Rightarrow {{x}^{3}}=\sqrt{288}\left( 288+\left[ 3\left( 119 \right) \right] \right)-\sqrt{119}\left( 119+\left[ 3\left( 288 \right) \right] \right) \\
& \Rightarrow {{x}^{3}}=\sqrt{288}\left( 288+\left[ 3\left( 119 \right) \right] \right)-\sqrt{119}\left( 119+\left[ 3\left( 288 \right) \right] \right) \\
& \Rightarrow {{x}^{3}}=\sqrt{288}\left( 645 \right)-\sqrt{119}\left( 983 \right)\text{ }.........\left( 3 \right) \\
\end{align}\]
Now, let us assume that the value of $\left( \sqrt{288}+\sqrt{119} \right)$ be y. Thus we have:
$y=\sqrt{288}+\sqrt{119}$
On cubing both sides we get:
${{y}^{3}}={{\left( \sqrt{288}+\sqrt{119} \right)}^{3}}$
Now, we will use another identity in above equation which is shown below:
${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}$
Thus, we will get:
${{y}^{3}}={{\left( \sqrt{288} \right)}^{3}}+{{\left( \sqrt{119} \right)}^{3}}+3{{\left( \sqrt{288} \right)}^{2}}\left( \sqrt{119} \right)+3\left( \sqrt{288} \right){{\left( \sqrt{119} \right)}^{2}}$
$\Rightarrow {{y}^{3}}={{\left( \sqrt{288} \right)}^{3}}+{{\left( \sqrt{119} \right)}^{3}}+\left[ 3\left( 288 \right)\left( \sqrt{119} \right) \right]+\left[ 3\left( \sqrt{288} \right)\left( 119 \right) \right]$
Now, we will rearrange the above equation and we will get:
\[\begin{align}
& \Rightarrow {{y}^{3}}={{\left( \sqrt{288} \right)}^{3}}+\left[ 3\left( \sqrt{288} \right)\left( 119 \right) \right]+{{\left( \sqrt{119} \right)}^{3}}+\left[ 3\left( 288 \right)\left( \sqrt{119} \right) \right] \\
& \Rightarrow {{y}^{3}}=\sqrt{288}\left( {{\left( \sqrt{288} \right)}^{2}}+\left[ 3\left( 119 \right) \right] \right)+\sqrt{119}\left( {{\left( \sqrt{119} \right)}^{2}}+\left[ 3\left( 288 \right) \right] \right) \\
\end{align}\]
Now, we will square the terms which has the power 2 and therefore we will get:
\[\begin{align}
& \Rightarrow {{y}^{3}}=\sqrt{288}\left( 288+\left[ 3\left( 119 \right) \right] \right)+\sqrt{119}\left( 119+\left[ 3\left( 288 \right) \right] \right) \\
& \Rightarrow {{y}^{3}}=\sqrt{288}\left( 288+\left[ 3\left( 119 \right) \right] \right)+\sqrt{119}\left( 119+\left[ 3\left( 288 \right) \right] \right) \\
& \Rightarrow {{y}^{3}}=\sqrt{288}\left( 645 \right)+\sqrt{119}\left( 983 \right)\text{ }.........\left( 4 \right) \\
\end{align}\]
Now, we will multiply the equation $\left( 3 \right)$ and $\left( 4 \right)$. After doing this, we will get:
${{x}^{3}}{{y}^{3}}=\left( \sqrt{288}\left( 645 \right)-\sqrt{119}\left( 983 \right) \right)\left( \sqrt{288}\left( 645 \right)+\sqrt{119}\left( 983 \right) \right)$
Here, we will use the identity on the RHS: $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
$\begin{align}
& \Rightarrow {{x}^{3}}{{y}^{3}}={{\left( \sqrt{288}\left( 645 \right) \right)}^{2}}-{{\left( \sqrt{119}\left( 983 \right) \right)}^{2}} \\
& \Rightarrow {{x}^{3}}{{y}^{3}}=288{{\left( 645 \right)}^{2}}-119{{\left( 983 \right)}^{2}} \\
\end{align}$
Now. we will expand 645 and 983
$\Rightarrow {{x}^{3}}{{y}^{3}}=288{{\left( 288+\left( 3\times 119 \right) \right)}^{2}}-119{{\left( 119+\left( 3\times 288 \right) \right)}^{2}}$
Now, we will apply the identity that is ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ :
\[\Rightarrow {{x}^{3}}{{y}^{3}}=288\left( {{\left( 288 \right)}^{2}}+{{\left( 3\times 119 \right)}^{2}}+2\left( 288 \right)\left( 3\times 119 \right) \right)-119\left( {{\left( 119 \right)}^{2}}+{{\left( 3\times 288 \right)}^{2}}+2\left( 3\times 288 \right)\left( 119 \right) \right)\]
Now, we will multiply 288 and 119 into the brackets:
\[\begin{align}
& \Rightarrow {{x}^{3}}{{y}^{3}}=\left( {{\left( 288 \right)}^{3}}+288{{\left( 3\times 119 \right)}^{2}}+2{{\left( 288 \right)}^{2}}\left( 3\times 119 \right) \right)-\left( {{\left( 119 \right)}^{3}}+119{{\left( 3\times 288 \right)}^{2}}+2\left( 3\times 288 \right){{\left( 119 \right)}^{2}} \right) \\
& \Rightarrow {{x}^{3}}{{y}^{3}}=\left( {{\left( 288 \right)}^{3}}+9\times 288\times {{\left( 119 \right)}^{2}}+6{{\left( 288 \right)}^{2}}119 \right)-\left( {{\left( 119 \right)}^{3}}+119\times 9\times {{\left( 288 \right)}^{2}}+6\times 288{{\left( 119 \right)}^{2}} \right) \\
\end{align}\]
Now, we will arrange the above terms:
\[\begin{align}
& \Rightarrow {{x}^{3}}{{y}^{3}}={{\left( 288 \right)}^{3}}-{{\left( 119 \right)}^{3}}+9\times 288\times {{\left( 119 \right)}^{2}}-6\times 288{{\left( 119 \right)}^{2}}+6{{\left( 288 \right)}^{2}}119-119\times 9\times {{\left( 288 \right)}^{2}} \\
& \Rightarrow {{x}^{3}}{{y}^{3}}={{\left( 288 \right)}^{3}}-{{\left( 119 \right)}^{3}}+288{{\left( 119 \right)}^{2}}\left( 9-6 \right)+{{\left( 288 \right)}^{2}}119\left( 6-9 \right) \\
& \Rightarrow {{x}^{3}}{{y}^{3}}={{\left( 288 \right)}^{3}}-{{\left( 119 \right)}^{3}}+3{{\left( 119 \right)}^{2}}288-3{{\left( 288 \right)}^{2}}119 \\
\end{align}\]
Now, we know that: ${{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}={{\left( a-b \right)}^{3}}$
Therefore: $\Rightarrow {{x}^{3}}{{y}^{3}}={{\left( 288-119 \right)}^{3}}\Rightarrow {{x}^{3}}{{y}^{3}}={{\left( 169 \right)}^{3}}\text{ }.......\text{ Equation 5}$
Now, we know that
$P={{\left( \sqrt{288}-\sqrt{119} \right)}^{\dfrac{3}{2}}}{{\left( \sqrt{288}+\sqrt{119} \right)}^{\dfrac{3}{2}}}$
$\Rightarrow P={{x}^{\dfrac{3}{2}}}{{y}^{\dfrac{3}{2}}}$
$\Rightarrow P={{\left( {{x}^{3}} \right)}^{\dfrac{1}{2}}}{{\left( {{y}^{3}} \right)}^{\dfrac{1}{2}}}$
Now, we will use another identity: ${{a}^{x}}\times {{b}^{x}}={{\left( ab \right)}^{x}}$
Thus, we will get:
$\Rightarrow P={{\left( {{x}^{3}}{{y}^{3}} \right)}^{\dfrac{1}{2}}}$
$\Rightarrow P={{\left( {{169}^{3}} \right)}^{\dfrac{1}{2}}}$ (from equation 5)
$\Rightarrow P={{\left[ {{\left( 13\times 13 \right)}^{3}} \right]}^{\dfrac{1}{2}}}$
$\Rightarrow P={{\left( 13\times 13 \right)}^{\dfrac{3}{2}}}$
$\Rightarrow P={{\left( 13 \right)}^{3}}=13\times 13\times 13$
$\Rightarrow P=2197$
Thus, value of ${{\left( \sqrt{288}-\sqrt{119} \right)}^{\dfrac{3}{2}}}{{\left( \sqrt{288}+\sqrt{119} \right)}^{\dfrac{3}{2}}}=2197$
Note: The above question can also be solved alternatively as shown below:
$P={{\left( \sqrt{288}-\sqrt{119} \right)}^{\dfrac{3}{2}}}{{\left( \sqrt{288}+\sqrt{119} \right)}^{\dfrac{3}{2}}}$
Here, we will use identity: ${{a}^{x}}\times {{b}^{x}}={{\left( ab \right)}^{x}}$
$\Rightarrow P={{\left[ \left( \sqrt{288}-\sqrt{119} \right)\left( \sqrt{288}+\sqrt{119} \right) \right]}^{\dfrac{3}{2}}}$
Now, we will use identity: $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$
$\Rightarrow P={{\left( 288-119 \right)}^{\dfrac{3}{2}}}$
$\Rightarrow P={{\left( 169 \right)}^{\dfrac{3}{2}}}$
$\Rightarrow P={{\left( 13\times 13 \right)}^{\dfrac{3}{2}}}$
$\Rightarrow P={{13}^{3}}$
$\Rightarrow P=2197$
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