Prove that \[{{\left( a+b \right)}^{5}}-{{a}^{5}}-{{b}^{5}}=5ab\left( a+b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]

Hint: We solve this problem first taking the LHS and expanding the power by using the binomial theorem. The binomial theorem states that
\[{{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}.{{x}^{n-r}}.{{y}^{r}}}\]
Where, \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
By using this binomial theorem we expand the term \[{{\left( a+b \right)}^{5}}\] and then rearrange the terms and take the common terms out to get the RHS which proves the required result.

Complete step by step answer:
We are asked to prove
\[\Rightarrow {{\left( a+b \right)}^{5}}-{{a}^{5}}-{{b}^{5}}=5ab\left( a+b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]
Let us take the LHS as follows
\[\Rightarrow LHS={{\left( a+b \right)}^{5}}-{{a}^{5}}-{{b}^{5}}......equation(i)\]
We know that the binomial theorem states that
\[{{\left( x+y \right)}^{n}}=\sum\limits_{r=0}^{n}{{}^{n}{{C}_{r}}.{{a}^{n-r}}.{{b}^{r}}}\]
By using the above formula let us find the value of \[{{\left( a+b \right)}^{5}}\] then we get’
\[\Rightarrow {{\left( a+b \right)}^{5}}=\sum\limits_{r=0}^{5}{{}^{5}{{C}_{r}}.{{a}^{5-r}}.{{b}^{r}}}\]

Now by expanding the summation to all terms we get
\[\Rightarrow {{\left( a+b \right)}^{5}}={}^{5}{{C}_{0}}.{{a}^{5-0}}.{{b}^{0}}+{}^{5}{{C}_{1}}.{{a}^{5-1}}.{{b}^{1}}+{}^{5}{{C}_{2}}.{{a}^{5-2}}.{{b}^{2}}+{}^{5}{{C}_{3}}.{{a}^{5-3}}.{{b}^{3}}+{}^{5}{{C}_{4}}.{{a}^{5-4}}.{{b}^{4}}+{}^{5}{{C}_{5}}.{{a}^{5-5}}.{{b}^{5}}\]
We know that the formula of combination as
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
By using the above formula we get
\[\Rightarrow {{\left( a+b \right)}^{5}}={{a}^{5}}+5{{a}^{4}}b+10{{a}^{3}}{{b}^{2}}+10{{a}^{2}}{{b}^{3}}+5a{{b}^{4}}+{{b}^{5}}\]
Now by substituting the value of \[{{\left( a+b \right)}^{5}}\] in equation (i) we get
\[\begin{align}
  & \Rightarrow LHS={{a}^{5}}+5{{a}^{4}}b+10{{a}^{3}}{{b}^{2}}+10{{a}^{2}}{{b}^{3}}+5a{{b}^{4}}+{{b}^{5}}-{{a}^{5}}-{{b}^{5}} \\
 & \Rightarrow LHS=5{{a}^{4}}b+10{{a}^{3}}{{b}^{2}}+10{{a}^{2}}{{b}^{3}}+5a{{b}^{4}} \\
\end{align}\]
Now, by taking the common terms out from the above equation we get
\[\Rightarrow LHS=5ab\left( {{a}^{3}}+{{b}^{3}}+2ab\left( a+b \right) \right)\]
We know that the formula of algebra that is
By using this formula in above equation we get
\[\begin{align}
  & \Rightarrow LHS=5ab\left[ \left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right)+2ab\left( a+b \right) \right] \\
 & \Rightarrow LHS=5ab\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}}+2ab \right) \\
 & \Rightarrow LHS=5ab\left( a+b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right) \\
\end{align}\]
Now let us take the RHS as
\[\begin{align}
  & \Rightarrow RHS=5ab\left( a+b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right) \\
 & \Rightarrow RHS=LHS \\
\end{align}\]
Hence the required result has been proved that is
\[{{\left( a+b \right)}^{5}}-{{a}^{5}}-{{b}^{5}}=5ab\left( a+b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]

Note: We can solve this problem in another way.
Here, we have the LHS as
\[\Rightarrow LHS=5{{a}^{4}}b+10{{a}^{3}}{{b}^{2}}+10{{a}^{2}}{{b}^{3}}+5a{{b}^{4}}\]
If we don’t get any idea of taking the common terms then leave the LHS as it is that is
\[\Rightarrow LHS=5{{a}^{4}}b+10{{a}^{3}}{{b}^{2}}+10{{a}^{2}}{{b}^{3}}+5a{{b}^{4}}\]
Now let us take the RHS as
\[\Rightarrow RHS=5ab\left( a+b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]
Now, by multiplying the terms in above equation we get
\[\Rightarrow RHS=5ab\left( {{a}^{3}}+{{a}^{2}}b+a{{b}^{2}}+{{a}^{2}}b+a{{b}^{2}}+{{b}^{3}} \right)\]
Now by taking the term \[5ab\] into the brackets and adding the same terms we get
\[\begin{align}
  & \Rightarrow RHS=5{{a}^{4}}b+10{{a}^{3}}{{b}^{2}}+10{{a}^{2}}{{b}^{3}}+5a{{b}^{4}} \\
 & \Rightarrow RHS=LHS \\
\end{align}\]
Hence the required result has been proved that is
\[{{\left( a+b \right)}^{5}}-{{a}^{5}}-{{b}^{5}}=5ab\left( a+b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]
By solving this method also the correct answer.