If we have algebraic expression as $a+b+c=0$ and ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1$ , then the value of ${{a}^{4}}+{{b}^{4}}+{{c}^{4}}$ is
(a) $\dfrac{1}{8}$
(b) $\dfrac{1}{4}$
(b) $\dfrac{1}{2}$
(d) $\dfrac{1}{3}$

Hint: In order to solve this problem, we need to know a few formulae. The main idea is to use the appropriate formulas and then substitute the necessary conditions as given in the question and arrive at the final answer. The necessary formulas are as follows, ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ac \right)$ , ${{\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}^{2}}={{a}^{4}}+{{b}^{4}}+{{c}^{4}}+2\left( {{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}} \right)$ , ${{\left( ab+bc+ac \right)}^{2}}={{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}}+2\left( {{a}^{2}}bc+a{{b}^{2}}c+ab{{c}^{2}} \right)$ .

Complete step-by-step solution:
We have given that $a+b+c=0$ and ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1$ .
We here need to find those identities where we could substitute these values and get the value of ${{a}^{4}}+{{b}^{4}}+{{c}^{4}}$.
Let’s start by solving for ${{\left( a+b+c \right)}^{2}}$ .
The identity says that,
${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ac \right)$
We know that LHS = 0 and ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1$ ,
Therefore, by substituting we get,
$0=1+2\left( ab+bc+ac \right)$
Solving this we get,
$\left( ab+bc+ac \right)=\dfrac{-1}{2}........................(i)$
Let’s start by solving for ${{\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}^{2}}$ .
The identity says that,
${{\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)}^{2}}={{a}^{4}}+{{b}^{4}}+{{c}^{4}}+2\left( {{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}} \right)$
Substituting the values of ${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=1$ , we get,
${{a}^{4}}+{{b}^{4}}+{{c}^{4}}+2\left( {{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}} \right)=1$
We need to find the value of ${{a}^{4}}+{{b}^{4}}+{{c}^{4}}$,
Hence, we get,
${{a}^{4}}+{{b}^{4}}+{{c}^{4}}=1-2\left( {{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}} \right)....................(ii)$
Like the previous identity, let's find the value of ${{\left( ab+bc+ac \right)}^{2}}$
The identity says that,
${{\left( ab+bc+ac \right)}^{2}}={{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}}+2\left( {{a}^{2}}bc+a{{b}^{2}}c+ab{{c}^{2}} \right)$
From equation (i), we get,
${{\left( \dfrac{-1}{2} \right)}^{2}}={{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}}+2\left( {{a}^{2}}bc+a{{b}^{2}}c+ab{{c}^{2}} \right)$
Solving this we get,
$\dfrac{1}{4}={{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}}+2abc\left( a+b+c \right)$
Also, from the initial condition, we know that $a+b+c=0$ .
Substituting we get,
$\begin{align}
  & \dfrac{1}{4}={{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}}+2abc\left( 0 \right) \\
 & {{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{a}^{2}}{{c}^{2}}=\dfrac{1}{4}...............(iii) \\
\end{align}$
Substituting the value of equation (iii) in equation (ii) we get,
${{a}^{4}}+{{b}^{4}}+{{c}^{4}}=1-2\left( \dfrac{1}{4} \right)$
Solving this we get,
${{a}^{4}}+{{b}^{4}}+{{c}^{4}}=1-\dfrac{1}{2}=\dfrac{1}{2}$ .

Hence the correct option is (b).

Note: In this problem, the idea behind using all the identities is the same. As we can see that we just need to understand only the first identity that is ${{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2\left( ab+bc+ac \right)$, and then we just have to substitute the values of a, b, c according to the requirement. Also, we can remember as the summation of individual squares of each term followed by all the possible combinations of the pairs multiplied by two.