Question

Find the cube root of the rational number $\dfrac{686}{-3456}$ .

Answer 1

Hint: To find the cube root of $\dfrac{686}{-3456}$ , we have to first write this mathematically as $\sqrt[3]{\dfrac{686}{-3456}}$ . Then, we will write -3456 as $-1\times 3456$ . We have to write -1 in terms of powers of 3. Then, we will apply the rules associated with roots, that is, $\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$ and $\sqrt[n]{a\times b}=\sqrt[n]{a}\times \sqrt[n]{b}$ . We will then find the cube roots of the result using the ladder method. The result is not a perfect cube, then we will write their prime factors and simplify.

Complete step by step answer:
We have to find the cube root of the rational number $\dfrac{686}{-3456}$ . Let us write this mathematically as
$\begin{align}
  & \Rightarrow \sqrt[3]{\dfrac{686}{-3456}} \\
 & =\sqrt[3]{\dfrac{686}{-1\times 3456}} \\
\end{align}$
We know that $\left( -1 \right)$ raised to an odd power is -1 itself. Therefore, we can write the above form as
$\Rightarrow \sqrt[3]{\dfrac{686}{-3456}}=\sqrt[3]{\dfrac{686}{{{\left( -1 \right)}^{3}}\times 3456}}$
We know that $\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$ . Therefore, the above result can be written as
$\Rightarrow \sqrt[3]{\dfrac{686}{-3456}}=\dfrac{\sqrt[3]{686}}{\sqrt[3]{{{\left( -1 \right)}^{3}}\times 3456}}$
We know that $\sqrt[n]{a\times b}=\sqrt[n]{a}\times \sqrt[n]{b}$ .
$\Rightarrow \sqrt[3]{\dfrac{686}{-3456}}=\dfrac{\sqrt[3]{686}}{\sqrt[3]{{{\left( -1 \right)}^{3}}}\times \sqrt[3]{3456}}$
Let us write the cube root of ${{\left( -1 \right)}^{3}}$ .
$\begin{align}
  & \Rightarrow \sqrt[3]{\dfrac{686}{-3456}}=\dfrac{\sqrt[3]{686}}{-1\times \sqrt[3]{3456}} \\
 & \Rightarrow \sqrt[3]{\dfrac{686}{-3456}}=\dfrac{\sqrt[3]{686}}{-\sqrt[3]{3456}}...\left( i \right) \\
\end{align}$
Let us first find the cube root of numerator and denominator.

Let us consider 686 and find its cube root using ladder method.
$\begin{align}
  & 2\left| \!{\underline {\,
  686 \,}} \right. \\
 & 7\left| \!{\underline {\,
  343 \,}} \right. \\
 & 7\left| \!{\underline {\,
  49 \,}} \right. \\
 & 7\left| \!{\underline {\,
  7 \,}} \right. \\
 & \text{ }\text{ }\text{ }1 \\
\end{align}$
We can see that 686 is not a perfect cube. So let us write the prime factors of 686 as $2\times 7\times 7\times 7=2\times {{7}^{3}}$
Now, let us find the cube root of 3456.
\[\begin{align}
  & 2\left| \!{\underline {\,
  3456 \,}} \right. \\
 & 2\left| \!{\underline {\,
  1728 \,}} \right. \\
 & 2\left| \!{\underline {\,
  864 \,}} \right. \\
 & 2\left| \!{\underline {\,
  432 \,}} \right. \\
 & 2\left| \!{\underline {\,
  216 \,}} \right. \\
 & 2\left| \!{\underline {\,
  108 \,}} \right. \\
 & 2\left| \!{\underline {\,
  54 \,}} \right. \\
 & 3\left| \!{\underline {\,
  27 \,}} \right. \\
 & 3\left| \!{\underline {\,
  9 \,}} \right. \\
 & 3\left| \!{\underline {\,
  3 \,}} \right. \\
 & \text{ }\text{ }\text{ }1 \\
\end{align}\]
We can write 3456 as $2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3$ . Let us try to group the same factors as a set of three.
$3456=\boxed{2\times 2\times 2}\times\boxed{ 2\times 2\times 2}\times 2\times\boxed{ 3\times 3\times 3}$
We cannot group all the prime factors. Therefore 3456 is not a perfect cube.
$3456=2\times 2\times 2\times 2\times 2\times 2\times 2\times 3\times 3\times 3={{2}^{7}}\times {{3}^{3}}$
Now, let us substitute the prime factors of 686 and 3456 in (i).
$\Rightarrow \sqrt[3]{\dfrac{686}{-3456}}=\dfrac{\sqrt[3]{2\times {{7}^{3}}}}{-\sqrt[3]{{{2}^{7}}\times {{3}^{3}}}}$
We know that $\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$ . Therefore, the above result can be written as
$\Rightarrow \sqrt[3]{\dfrac{686}{-3456}}=-\sqrt[3]{\dfrac{2\times {{7}^{3}}}{{{2}^{7}}\times {{3}^{3}}}}$
Let us cancel 2 from the numerator and the denominator.
$\Rightarrow \sqrt[3]{\dfrac{686}{-3456}}=-\sqrt[3]{\dfrac{\require{cancel}\cancel{2}\times {{7}^{3}}}{{{2}^{{{\require{cancel}\cancel{7}}^{6}}}}\times {{3}^{3}}}}$
We can write the result of the above simplification as
$\Rightarrow \sqrt[3]{\dfrac{686}{-3456}}=-\sqrt[3]{\dfrac{{{7}^{3}}}{{{2}^{6}}\times {{3}^{3}}}}$
Let us write ${{2}^{6}}$ as ${{2}^{3}}\times {{2}^{3}}$ since ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ .
$\Rightarrow \sqrt[3]{\dfrac{686}{-3456}}=-\sqrt[3]{\dfrac{{{7}^{3}}}{{{2}^{3}}\times {{2}^{3}}\times {{3}^{3}}}}$
We know that $\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}$ . Therefore, the above result can be written as
$\Rightarrow \sqrt[3]{\dfrac{686}{-3456}}=-\dfrac{\sqrt[3]{{{7}^{3}}}}{\sqrt[3]{{{2}^{3}}\times {{2}^{3}}\times {{3}^{3}}}}$
We know that $\sqrt[n]{a\times b}=\sqrt[n]{a}\times \sqrt[n]{b}$ .
$\Rightarrow \sqrt[3]{\dfrac{686}{-3456}}=-\dfrac{\sqrt[3]{{{7}^{3}}}}{\sqrt[3]{{{2}^{3}}}\times \sqrt[3]{{{2}^{3}}}\times \sqrt[3]{{{3}^{3}}}}$
We can write the result of the above equation as
$\begin{align}
  & \Rightarrow \sqrt[3]{\dfrac{686}{-3456}}=-\dfrac{7}{2\times 2\times 3} \\
 & \Rightarrow \sqrt[3]{\dfrac{686}{-3456}}=-\dfrac{7}{12} \\
\end{align}$
Hence, the cube root of the rational number $\dfrac{686}{-3456}$ is $-\dfrac{7}{12}$ .

Note: Students must be thorough with the rules associated with roots. They must know to find the n root of the numbers, where $n=1,2,3,...$ . They must understand thoroughly the rules of exponents. Students must never miss to take the cube root of -1 in the denominator.