Simplify the following expression:
$\dfrac{{{m}^{2}}-{{n}^{2}}}{{{\left( m+n \right)}^{2}}}\times \dfrac{{{m}^{2}}+mn+{{n}^{2}}}{{{m}^{3}}-{{n}^{3}}}$

Hint: To simplify the given expression, we are going to use the following algebraic identities. We know that $\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$, \[\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]. You can see that in the expression given above $\left( {{m}^{2}}-{{n}^{2}} \right)\And \left( {{m}^{3}}-{{n}^{3}} \right)$ are in the form of the algebraic identities that we have just shown. After applying the identities in the expression using simple algebra we can simplify it.

Complete step-by-step solution:
We have given the following expression:
$\dfrac{{{m}^{2}}-{{n}^{2}}}{{{\left( m+n \right)}^{2}}}\times \dfrac{{{m}^{2}}+mn+{{n}^{2}}}{{{m}^{3}}-{{n}^{3}}}$……………...…. Eq. (1)
We have to simplify the above expression.
If you carefully look at the given expression you will find that $\left( {{m}^{2}}-{{n}^{2}} \right)$ is given and it is written in the form of $\left( {{a}^{2}}-{{b}^{2}} \right)$ and we know that the identity of $\left( {{a}^{2}}-{{b}^{2}} \right)$ which is equal to:
$\left( {{a}^{2}}-{{b}^{2}} \right)=\left( a+b \right)\left( a-b \right)$
So, we can write $\left( {{m}^{2}}-{{n}^{2}} \right)$ as $\left( m+n \right)\left( m-n \right)$ in the given expression.
$\dfrac{\left( m+n \right)\left( m-n \right)}{{{\left( m+n \right)}^{2}}}\times \dfrac{{{m}^{2}}+mn+{{n}^{2}}}{{{m}^{3}}-{{n}^{3}}}$……………...…. Eq. (2)
Now, the term $\left( {{m}^{3}}-{{n}^{3}} \right)$ is written in the form of $\left( {{a}^{3}}-{{b}^{3}} \right)$ and we know that the identity of $\left( {{a}^{3}}-{{b}^{3}} \right)$ is equal to:
$\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$
So, we can write $\left( {{m}^{3}}-{{n}^{3}} \right)$ as $\left( m-n \right)\left( {{m}^{2}}+mn+{{n}^{2}} \right)$ in eq. (2).
$\dfrac{\left( m+n \right)\left( m-n \right)}{{{\left( m+n \right)}^{2}}}\times \dfrac{{{m}^{2}}+mn+{{n}^{2}}}{\left( m-n \right)\left( {{m}^{2}}+mn+{{n}^{2}} \right)}$
Now, in the above expression $\left( m-n \right)$ will be cancelled out from the numerator and the denominator and we are left with:
$\dfrac{\left( m+n \right)}{{{\left( m+n \right)}^{2}}}\times \dfrac{{{m}^{2}}+mn+{{n}^{2}}}{\left( {{m}^{2}}+mn+{{n}^{2}} \right)}$
It is clearly visible from the above expression that $\left( {{m}^{2}}+mn+{{n}^{2}} \right)$ is cancelling out and the remaining expression will look like:
$\dfrac{\left( m+n \right)}{{{\left( m+n \right)}^{2}}}$
Now, in the above expression one $\left( m+n \right)$ will be cancelled out from the numerator and denominator and we are left with the expression as:
$\dfrac{1}{m+n}$
Hence, we have simplified the above expression to $\dfrac{1}{m+n}$.

Note: The most common mistake that could be possible in this problem is in wrongly writing the identity of \[\left( {{a}^{3}}-{{b}^{3}} \right)\] sometimes students considered it as \[\left(a-b \right)^{3}\] which makes the solution wrong. The correct identity of \[\left( {{a}^{3}}-{{b}^{3}} \right)\] is equal to:
\[\left( {{a}^{3}}-{{b}^{3}} \right)=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]
The mistake that could happen in writing the above identity is in putting the positive and negative signs.