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Last updated date: 13th Jun 2024
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Answer
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Hint: We apply the concept of cube root using the prime factorisation theorem. We break the main number into multiplications of prime. Then depending on the cube root of 343 we take one prime out of triplets of the same prime. At the end we verify it with the help of indices.

Complete step-by-step solution:
Let’s assume that the cube root of the number 343 is x. This means cube of x will be 343.
So, ${{x}^{3}}=343$ which gives $x=\sqrt[3]{343}={{343}^{\dfrac{1}{3}}}$.
Now we find the prime factorisation of the number 343.
$\begin{align}
  & 7\left| \!{\underline {\,
  343 \,}} \right. \\
 & 7\left| \!{\underline {\,
  49 \,}} \right. \\
 & 7\left| \!{\underline {\,
  7 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}$
So, $343={{7}^{3}}$. In case of finding the root, we take the common numbers out in order of triplets. This means when we find the square roots, we will take two same primes of the factorisations and treat them as only one. When we find cube roots, we will take three same primes of the factorisations and treat them as only one.
In case of 343, we have three 7s. At the time of taking cube root, we take only one 7 out of three.
So, $x=\sqrt[3]{343}=\sqrt[3]{7\times 7\times 7}=7$.
Therefore, the value of the cube root of the number 343 is 7.

Note: We can solve it using the law of indices. We know that \[{{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}}\]. Now here we need to find the value of $\sqrt[3]{343}={{343}^{\dfrac{1}{3}}}$. We know that $343={{7}^{3}}$.
So, we get $\sqrt[3]{343}={{343}^{\dfrac{1}{3}}}={{\left( {{7}^{3}} \right)}^{\dfrac{1}{3}}}={{7}^{\dfrac{3}{3}}}=7$.