Prove the following expression–
${{\left( a+b+c \right)}^{3}}-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}=3(a+b)(b+c)(c+a)$

Hint: To solve this we should know about the basics of expansion of algebraic expression. In this case, we should know how to expand an algebraic expression raised to the power of 3. The formula is given by ${{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)$. We will make use of this fact to solve this problem.

Complete step-by-step answer:
In this problem, we need to prove that LHS and RHS are the same. Thus, to do so, we will start by evaluating the LHS. In particular, we will evaluate ${{(a+b+c)}^{3}}$ term of the LHS first. We have,
= ${{(a+b+c)}^{3}}$
= ${{((a+b)+c)}^{3}}$
Now, we can use the formula of ${{(a+b)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab(a+b)$, thus, we get,
= ${{(a+b)}^{3}}+{{c}^{3}}+3(a+b)c((a+b)+c)$
Now, we evaluate ${{(a+b)}^{3}}$, thus, we get,
= ${{a}^{3}}+{{b}^{3}}+3ab(a+b)+{{c}^{3}}+3(a+b)c((a+b)+c)$
= ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3ab(a+b)+3(a+b)c((a+b)+c)$
Now, taking 3(a+b) common, we have,
= ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3(a+b)[ab+ac+bc+{{c}^{2}}]$
= ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3(a+b)[a(b+c)+c(b+c)]$
= ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3(a+b)[(a+c)(b+c)]$ -- (1)

Now, we can finally evaluate the LHS term completely. Thus, we get,
= ${{\left( a+b+c \right)}^{3}}-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}$
Now, we also know the value of ${{(a+b+c)}^{3}}$ from (1). We substitute this in the above expression, we get,
= ${{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3(a+b)[(a+c)(b+c)]-{{a}^{3}}-{{b}^{3}}-{{c}^{3}}$
 = 3(a+b) (b+c) (a+c)
Clearly, this is the same as the RHS term. Thus, LHS=RHS. Hence, proved.

Note: An alternative way to proceed with this problem is to solve the RHS term instead of the LHS term. The RHS term is $3(a+b)(b+c)(c+a)$, thus, we can evaluate this and we would still be able to arrive at the LHS term. Although, this will be a more difficult task since, one should know the formula of ${{(a+b+c)}^{3}}$, which is not intuitive.