How do you simplify \[\sqrt{8}\left( \sqrt{13}-\sqrt{117} \right)\]?

Hint: In this problem, we have to simplify the given square root. We can first take the square root \[\sqrt{117}\] , we can separate it using the multiplication table and we can take the common terms outside. We can then simplify the terms inside the brackets and we can multiply the result with the terms in the square root to get the final simplified form.

Complete step by step answer:
We know that the given square root to be simplified is,
\[\sqrt{8}\left( \sqrt{13}-\sqrt{117} \right)\]
We can first take the square root \[\sqrt{117}\] , we can separate it using the multiplication table, we get
\[\Rightarrow \sqrt{8}\left( \sqrt{13}-\sqrt{13\times {{3}^{2}}} \right)\]
We can now simplify the terms, we get
\[\Rightarrow \sqrt{8}\left( \sqrt{13}-3\sqrt{13} \right)\]
Now we can take the common term outside the brackets, we get
\[\Rightarrow \sqrt{8}\sqrt{13}\left( 1-3 \right)\]
We can now take the square root \[\sqrt{8}\] , we can separate it using the multiplication table, we get
\[\Rightarrow \sqrt{{{2}^{2}}\times 2}\sqrt{13}\left( -2 \right)\]
We can now separate the terms and cancel the square and the square root, we get
\[\Rightarrow 2\sqrt{2}\sqrt{13}\left( -2 \right)\]
 We can now multiply the term in the bracket to the terms outside, we get
\[\Rightarrow -4\sqrt{2}\sqrt{13}\]
We can now combine the above square root terms using multiplication of square roots.
We know that the multiplication of roots formula is,
\[\Rightarrow \sqrt{xy}=\sqrt{x}\times \sqrt{y}\]
We can apply this formula in the above step, we get
\[\Rightarrow -4\sqrt{2\times 13}=-4\sqrt{26}\]
Therefore, the simplified form of \[\sqrt{8}\left( \sqrt{13}-\sqrt{117} \right)\] is \[-4\sqrt{26}\].

Note:
Students make mistakes while splitting the terms according to the similar term which can be taken as a common term for simplifying the given square root in an easier way. We should also concentrate on the multiplication of square roots formula, which is used in these types of problems.