Evaluate \[\sqrt[3]{121}\times \sqrt[3]{297}\]

Hint: Type of questions are based on the algebraic expression which can be easily solved if we know some of the basic algebraic identities, and how to use them. Further you should also try to make it as simple as these questions, through converting them into prime numbers or simple numbers. Here are some of the identities which we will required in this question, to simply it are;
\[\begin{align}
  & {{a}^{n}}{{a}^{m}}={{a}^{(n+m)}} \\
 & {{a}^{\left( \dfrac{1}{n} \right)}}=\sqrt[n]{a} \\
 & {{(ab)}^{m}}={{a}^{m}}{{b}^{m}} \\
 & {{({{a}^{n}})}^{m}}={{a}^{mn}} \\
\end{align}\]
In these identities a, b, m, n can be variable or constant. We will use these identities to solve the question.

Complete step by step answer:
Moving further with the question, i.e. \[\sqrt[3]{121}\times \sqrt[3]{297}\]
Now you should aim to reduce these large values into prime numbers and try to vanish these exponential values using the above identities.
So by using the second identity i.e. \[{{a}^{\left( \dfrac{1}{n} \right)}}=\sqrt[n]{a}\]
${{\left( 121 \right)}^{\dfrac{1}{3}}}\times {{\left( 297 \right)}^{\dfrac{1}{3}}}$
To reduce the numbers to prime numbers, prime factorisation of 121 and 297 is;
$\begin{align}
  & 121=11\times 11 \\
 & 297=11\times 3\times 3\times 3 \\
\end{align}$
So,${{\left( 121 \right)}^{\dfrac{1}{3}}}\times {{\left( 297 \right)}^{\dfrac{1}{3}}}$can be further written as in prime numbers;
 \[\begin{align}
  & {{(11\times 11)}^{\left( \dfrac{1}{3} \right)}}\times {{(11\times 3\times 3\times 3)}^{\left( \dfrac{1}{3} \right)}} \\
 & {{({{11}^{1+1}})}^{\left( \dfrac{1}{3} \right)}}\times {{({{11}^{1}}\times {{3}^{1+1+1}})}^{\left( \dfrac{1}{3} \right)}} \\
 & {{({{11}^{2}})}^{\left( \dfrac{1}{3} \right)}}\times {{({{11}^{1}}\times {{3}^{3}})}^{\left( \dfrac{1}{3} \right)}} \\
\end{align}\]
By using the third identity i.e. \[{{(ab)}^{m}}={{a}^{m}}{{b}^{m}}\]we can write above equation as
\[\begin{align}
  & {{({{11}^{2}})}^{\left( \dfrac{1}{3} \right)}}\times {{({{11}^{1}}\times {{3}^{3}})}^{\left( \dfrac{1}{3} \right)}} \\
 & {{({{11}^{2}})}^{\left( \dfrac{1}{3} \right)}}\times {{(11)}^{\left( \dfrac{1}{3} \right)}}\times {{({{3}^{3}})}^{\left( \dfrac{1}{3} \right)}} \\
\end{align}\]
Further, by using the fourth identity i.e. \[{{(\mathop{a}^{n})}^{m}}={{a}^{mn}}\], we can write above expression as
\[\begin{align}
  & {{({{11}^{2}})}^{\left( \dfrac{1}{3} \right)}}\times {{(11)}^{\left( \dfrac{1}{3} \right)}}\times {{({{3}^{3}})}^{\left( \dfrac{1}{3} \right)}} \\
 & {{(11)}^{\left( \dfrac{2}{3} \right)}}\times {{(11)}^{\left( \dfrac{1}{3} \right)}}\times {{(3)}^{\left( \dfrac{3}{3} \right)}} \\
\end{align}\]
And finally by using the first identity i.e. \[{{a}^{n}}{{a}^{m}}={{a}^{(n+m)}}\], we can write above equation as;
\[\begin{align}
  & {{(11)}^{\left( \dfrac{2}{3} \right)}}\times {{(11)}^{\left( \dfrac{1}{3} \right)}}\times {{(3)}^{\left( \dfrac{3}{3} \right)}} \\
 & {{(11)}^{\left( \dfrac{2}{3}+\dfrac{1}{3} \right)}}\times {{(3)}^{\left( \dfrac{3}{3} \right)}} \\
 & {{(11)}^{\left( \dfrac{3}{3} \right)}}\times {{(3)}^{\left( \dfrac{3}{3} \right)}} \\
 & {{(11)}^{\left( 1 \right)}}\times {{(3)}^{\left( 1 \right)}} \\
\end{align}\]
So, we get $11\times 3$ on solving, which is 33.
Hence 33 is the answer.

Note: while using identities keep in mind that power gets added only when two terms are same and in multiplication. And whenever you try to simplify such a question, always try to reduce these large values into smaller one, and prime numbers are the smaller one.