If y is the mean proportional between x and z, Prove that $xyz{{\left( x+y+z \right)}^{3}}={{\left( xy+yz+zx \right)}^{3}}$.

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Hint: We start solving this question by using the definition of mean proportional in the starting which is, mean proportional of two numbers a and b is $\sqrt{ab}$. Then we apply the basic exponential formulas ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ and ${{a}^{m}}\times {{b}^{m}}={{\left( ab \right)}^{m}}$.Using these we convert the given expression to prove the given statement.

Complete step by step answer:
First, let us consider the definition of mean proportional.
Mean Proportional of any two numbers is the square root of the product of the two numbers.
So, from the definition the mean proportional of x and z is $\sqrt{xz}$.
We were given that y is the mean proportional between x and z. So, we can write y as,
$\Rightarrow y=\sqrt{xz}$
Now, let us consider the expression, $xyz{{\left( x+y+z \right)}^{3}}$.
As we know that the value of y is $\sqrt{xz}$. We substitute the value of y in the above expression.
\[\begin{align}
  & \Rightarrow xyz{{\left( x+y+z \right)}^{3}}..................\left( 1 \right) \\
 & \Rightarrow x\left( \sqrt{xz} \right)z{{\left( x+\sqrt{xz}+z \right)}^{3}} \\
 & \Rightarrow xz\left( \sqrt{xz} \right){{\left( x+\sqrt{xz}+z \right)}^{3}} \\
\end{align}\]
Let us use the formula $a={{\left( \sqrt{a} \right)}^{2}}$.
Then we can write $xz$ as ${{\left( \sqrt{xz} \right)}^{2}}$
\[\Rightarrow {{\left( \sqrt{xz} \right)}^{2}}\left( \sqrt{xz} \right){{\left( x+\sqrt{xz}+z \right)}^{3}}.............\left( 2 \right)\]
Now, we use the formula ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$. So,
$\begin{align}
  & \Rightarrow {{\left( \sqrt{xz} \right)}^{2}}\times {{\left( \sqrt{xz} \right)}^{1}}={{\left( \sqrt{xz} \right)}^{2+1}} \\
 & \Rightarrow {{\left( \sqrt{xz} \right)}^{2}}\times {{\left( \sqrt{xz} \right)}^{1}}={{\left( \sqrt{xz} \right)}^{3}} \\
\end{align}$
Using this and substituting in equation (2), we get,
\[\Rightarrow {{\left( \sqrt{xz} \right)}^{3}}{{\left( x+\sqrt{xz}+z \right)}^{3}}\]
Now, we use the formula ${{a}^{m}}\times {{b}^{m}}={{\left( ab \right)}^{m}}$. So, using this formula we get,
\[\begin{align}
  & \Rightarrow {{\left( x\sqrt{xz}+\sqrt{xz}\sqrt{xz}+z\sqrt{xz} \right)}^{3}} \\
 & \Rightarrow {{\left( x\sqrt{xz}+xz+z\sqrt{xz} \right)}^{3}} \\
\end{align}\]
As we know that $y=\sqrt{xz}$,
\[\begin{align}
  & \Rightarrow {{\left( x\sqrt{xz}+xz+z\sqrt{xz} \right)}^{3}} \\
 & \Rightarrow {{\left( xy+xz+zy \right)}^{3}} \\
 & \Rightarrow {{\left( xy+yz+zx \right)}^{3}}.............\left( 3 \right) \\
\end{align}\]
So, by using equations (1) and (3), we can write that
$\Rightarrow xyz{{\left( x+y+z \right)}^{3}}={{\left( xy+yz+zx \right)}^{3}}$
Hence Proved.

Note: There is a chance of making a mistake while using the formulas by taking ${{a}^{m}}\times {{b}^{m}}={{\left( ab \right)}^{m+m}}={{\left( ab \right)}^{2m}}$ confusing it with the formula ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$. But it is wrong because we should add the powers only when the bases are equal in the product. So, as the bases here are different, we should use ${{a}^{m}}\times {{b}^{m}}={{\left( ab \right)}^{m}}$. So, one should remember the formulas of exponential correctly.