1. Simplify the following: $\sqrt{72}+\sqrt{800}-\sqrt{18}$
2. Simplify: ${{343}^{-\dfrac{1}{3}}}$

Hint: For answering this question we need to simplify the 2 expressions which we have from the question and extract values by using exponents. We can use the formulae ${{\left( {{a}^{n}} \right)}^{\dfrac{1}{n}}}=\sqrt[n]{{{a}^{n}}}=\left( a \right)$ from the basic concept of exponents.

Complete step-by-step solution:
Now considering from the question we have the expression $\sqrt{72}+\sqrt{800}-\sqrt{18}$.
This expression can be simplified as $\sqrt{{{2}^{3}}\times {{3}^{2}}}+\sqrt{800}-\sqrt{18}$ .
And we can also more simplify the expression and write it as $\sqrt{{{2}^{3}}\times {{3}^{2}}}+\sqrt{800}-\sqrt{2\times {{3}^{2}}}$
This expression can be further simplified and written as $\sqrt{{{2}^{3}}\times {{3}^{2}}}+\sqrt{{{2}^{3}}\times {{10}^{2}}}-\sqrt{2\times {{3}^{2}}}$ .
For further simplifying we will use the basic concept of exponents ${{\left( {{a}^{n}} \right)}^{\dfrac{1}{n}}}=\sqrt[n]{{{a}^{n}}}=\left( a \right)$ and then we will have as follows.
This can be further simplified as $3\left( 2 \right)\sqrt{2}+10\left( 2 \right)\sqrt{2}-3\sqrt{2}$.
By simplifying this expression we will have
 $\begin{align}
  & \Rightarrow 6\sqrt{2}+20\sqrt{2}-3\sqrt{2} \\
 & =23\sqrt{2} \\
\end{align}$
Therefore we can conclude that $\sqrt{72}+\sqrt{800}-\sqrt{18}=23\sqrt{2}$ .
2. Now considering from the question we have the expression ${{343}^{-\dfrac{1}{3}}}$.
This expression can be simplified and written as ${{\left( {{7}^{3}} \right)}^{-\dfrac{1}{3}}}$ .
This can be further simplified as ${{\left( {{7}^{3}} \right)}^{-\dfrac{1}{3}}}$.
For further simplifying we will use the basic concept of exponents ${{\left( {{a}^{n}} \right)}^{\dfrac{1}{n}}}=\sqrt[n]{{{a}^{n}}}=\left( a \right)$ and then we will have as follows.
By performing further simplifications we will have ${{7}^{-1}}=\dfrac{1}{7}$.
Therefore we can conclude that ${{343}^{-\dfrac{1}{3}}}=\dfrac{1}{7}$.
Hence we can say that the simplified values of the expressions are $\sqrt{72}+\sqrt{800}-\sqrt{18}=23\sqrt{2}$ and ${{343}^{-\dfrac{1}{3}}}=\dfrac{1}{7}$.

Note: While answering questions of this type we have to be sure with the calculations. From the basic concept of exponents we will use the basic formulae saying that ${{\left( {{a}^{n}} \right)}^{\dfrac{1}{n}}}=\sqrt[n]{{{a}^{n}}}=\left( a \right)$. It would be efficient for answering questions of this type if we remember the squares and cubes of the first 10 natural numbers.