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Find the cube root of $38+17\sqrt{5}$.

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Last updated date: 22nd Jul 2024
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Answer
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Hint: For an irrational number, the irrational part i.e. the number inside the square root is the same for the number and it’s cube root. So, we can assume the cube root of $38+17\sqrt{5}$ as $a+b\sqrt{5}$ .

Before proceeding with the question, we must know that for an irrational number, the irrational part i.e. the number inside the square root is the same for the number and it’s cube root.
In the question, we are asked to find the cube root of an irrational number which is $38+17\sqrt{5}$. Since the irrational part i.e. the number inside the square root is the same for the number and it’s cube root, we can assume the cube root of $38+17\sqrt{5}$ as $a+b\sqrt{5}$.
As $a+b\sqrt{5}$ is the cube root of $38+17\sqrt{5}$, we can say that the cube of $a+b\sqrt{5}$ is equal to $38+17\sqrt{5}$. Hence, we can write,
\[{{\left( a+b\sqrt{5} \right)}^{3}}=38+17\sqrt{5}............\left( 1 \right)\]
Also, we have a formula which can be used to solve the above equation. The formula is,
${{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3{{x}^{2}}y+3x{{y}^{2}}................\left( 2 \right)$
From equation $\left( 1 \right)$, substituting $x=a$ and $y=b\sqrt{5}$ in equation $\left( 2 \right)$, we get.
  \[\begin{align}
  & {{\left( a+b\sqrt{5} \right)}^{3}}={{a}^{3}}+{{\left( b\sqrt{5} \right)}^{3}}+3{{a}^{2}}\left( b\sqrt{5} \right)+3a{{\left( b\sqrt{5} \right)}^{2}} \\
 & \Rightarrow {{\left( a+b\sqrt{5} \right)}^{3}}={{a}^{3}}+5{{b}^{3}}\sqrt{5}+3{{a}^{2}}b\sqrt{5}+15a{{b}^{2}} \\
 & \Rightarrow {{\left( a+b\sqrt{5} \right)}^{3}}={{a}^{3}}+15a{{b}^{2}}+5{{b}^{3}}\sqrt{5}+3{{a}^{2}}b\sqrt{5} \\
 & \Rightarrow {{\left( a+b\sqrt{5} \right)}^{3}}={{a}^{3}}+15a{{b}^{2}}+\sqrt{5}\left( 5{{b}^{3}}+3{{a}^{2}}b \right)..........\left( 3 \right) \\
\end{align}\]
Substituting \[{{\left( a+b\sqrt{5} \right)}^{3}}={{a}^{3}}+15a{{b}^{2}}+\sqrt{5}\left( 5{{b}^{3}}+3{{a}^{2}}b \right)\] from equation $\left( 3 \right)$ in equation $\left( 1 \right)$, we get,
\[{{a}^{3}}+15a{{b}^{2}}+\sqrt{5}\left( 5{{b}^{3}}+3{{a}^{2}}b \right)=38+17\sqrt{5}...........\left( 4 \right)\]
Comparing rational and irrational parts on both the sides of the equation, we get,
   \[{{a}^{3}}+15a{{b}^{2}}=38...........\left( 5 \right)\]
\[5{{b}^{3}}+3{{a}^{2}}b=17..........\left( 6 \right)\]
Form equation $\left( 5 \right)$, we have,
\[\begin{align}
  & 15a{{b}^{2}}=38-{{a}^{3}} \\
 & \Rightarrow {{b}^{2}}=\dfrac{38-{{a}^{3}}}{15a}.............\left( 7 \right) \\
\end{align}\]
Form equation $\left( 6 \right)$, we have,
\[b\left( 5{{b}^{2}}+3{{a}^{2}} \right)=17\]
Substituting \[{{b}^{2}}=\dfrac{38-{{a}^{3}}}{15a}\] in the above equation, we have,
\[\begin{align}
  & b\left( 5.\dfrac{38-{{a}^{3}}}{15a}+3{{a}^{2}} \right)=17 \\
 & \Rightarrow b\left( \dfrac{38-{{a}^{3}}}{3a}+3{{a}^{2}} \right)=17 \\
 & \Rightarrow b\left( \dfrac{38-{{a}^{3}}+9{{a}^{3}}}{3a} \right)=17 \\
 & \Rightarrow b\left( \dfrac{38+8{{a}^{3}}}{3a} \right)=17 \\
 & \Rightarrow b=\dfrac{51a}{38+8{{a}^{3}}} \\
\end{align}\]
Substituting $b$ from the above equation in equation $\left( 7 \right)$, we get,
$\begin{align}
  & {{\left( \dfrac{51a}{38+8{{a}^{3}}} \right)}^{2}}=\dfrac{38-{{a}^{3}}}{15a} \\
 & \Rightarrow {{51}^{2}}.15{{a}^{3}}=\left( 38-{{a}^{3}} \right){{\left( 38+8{{a}^{3}} \right)}^{2}} \\
\end{align}$
Let us substitute ${{a}^{3}}=x$ in the above equation.
$\Rightarrow {{51}^{2}}.15x=\left( 38-x \right){{\left( 38+8x \right)}^{2}}$
Since the above equation in variable $x$ is an equation of degree $3$, we can solve it only by hit and trial method. This means we have to randomly consider different integral values for $x$ and then we have to check which integral value of $x$ is satisfying this equation.
By hit and trial method, we get $x=8$. Since $x={{a}^{3}}$, we can say that ${{a}^{3}}=8$. Hence, $a=2$.
Substituting $a=2$ in equation $\left( 7 \right)$, we get,
\[\begin{align}
  & {{b}^{2}}=\dfrac{38-{{2}^{3}}}{15.2} \\
 & \Rightarrow {{b}^{2}}=\dfrac{38-8}{30} \\
 & \Rightarrow {{b}^{2}}=\dfrac{30}{30} \\
 & \Rightarrow {{b}^{2}}=1 \\
 & \Rightarrow b=1,b=-1 \\
\end{align}\]
For $b=-1$, $a+b\sqrt{5}$ will become negative which is not possible since $a+b\sqrt{5}$ is a cube root of a positive number.
Hence, $b=1$.
So, we get $a=2,b=1$.
Hence the cube root of $38+17\sqrt{5}$ is equal to $2+\sqrt{5}$.

Note: There is an alternative method to solve this question. We can write $38+17\sqrt{5}$ as
$8+5\sqrt{5}+12\sqrt{5}+30$ which can be again written as ${{2}^{3}}+{{\left( \sqrt{5} \right)}^{3}}+3{{\left( 2 \right)}^{2}}\left( \sqrt{5} \right)+3\left( 2 \right){{\left( \sqrt{5} \right)}^{2}}$. If we notice this expression carefully, we will find that it is an expansion of ${{\left( 2+\sqrt{5} \right)}^{3}}$. Hence, we get $38+17\sqrt{5}={{\left( 2+\sqrt{5} \right)}^{3}}$. This means that $\left( 2+\sqrt{5} \right)$ is the cube root of $38+17\sqrt{5}$.