Consider the given expression $i=\sqrt{-1}$ , then the value of the given expression $4+5{{\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}$ is equal to
(a) $1-i\sqrt{3}$
(b) $-1+i\sqrt{3}$
(c) $i\sqrt{3}$
(d) $-i\sqrt{3}$

Hint: Convert the imaginary number which is in the terms of “i” into an expression in terms of “ $\omega $ ”. And then solve the equation. We know $\omega $ is a root of the equation: ${{x}^{3}}=1$
Finding the value of $\omega $
${{x}^{3}}-1=0..........(A)$
So, we need formula of ${{x}^{n}}-{{y}^{n}}$

Complete step-by-step solution -
We need to prove:
${{a}^{n}}-{{b}^{n}}=\left( a-b \right)\left( {}^{n}{{C}_{1}}{{b}^{n-1}}+........{}^{n}{{C}_{n}}{{\left( a-b \right)}^{n-1}} \right)$
Proof of ${{a}^{n}}-{{b}^{n}}=\left( a-b \right)\left( {}^{n}{{C}_{1}}{{b}^{n-1}}+...........+{}^{n}{{C}_{n}}{{\left( a-b \right)}^{n-1}} \right)...........(i)$
By Binomial theorem, we get:
${{\left( a+b \right)}^{n}}=\sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{a}^{k}}{{b}^{n-k}}}$
By general algebraic knowledge, we can write “a” as:
$a=\left[ \left( a-b \right)+b \right]$
Putting this in left hand side of equation (i), we get:
${{a}^{n}}-{{b}^{n}}={{\left[ \left( a-b \right)+b \right]}^{n}}-{{b}^{n}}$
By using binomial theorem in this equation, we get:
${{a}^{n}}-{{b}^{n}}=\left( \sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}{{\left( a-b \right)}^{k}}{{b}^{n-k}}-{{b}^{n}}} \right)$
At k = 0 you can see a term ${{b}^{n}}$ in the summation term which gets cancelled by the last term in the equation.
By cancelling the common terms, we get:
${{a}^{n}}-{{b}^{n}}=\left( \sum\limits_{k=1}^{n}{{}^{n}{{C}_{k}}{{\left( a-b \right)}^{k}}{{b}^{n-k}}} \right)$
Now all terms are divisible by $\left( a-b \right)$
So, here $\left( a-b \right)$ is a common factor.
By taking the term $\left( a-b \right)$ common from expression we get:
${{a}^{n}}-{{b}^{n}}=\left( a-b \right)\left( {}^{n}{{C}_{1}}{{b}^{n-1}}+...........+{}^{n}{{C}_{n}}{{\left( a-b \right)}^{n-1}} \right)$
Now if we assume
$a=x,b=1,n=3$
By substituting the above, we get
${{x}^{3}}-1=\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)$
By substituting this into equation (A), we get:
$\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)=0$
The solutions of this equation:
$\begin{align}
  & x-1=0 \\
 & \Rightarrow x=1 \\
 & {{x}^{2}}+x+1=0 \\
\end{align}$
So, ${{x}^{2}}+x+1=0$
We need the solutions of the above expression.
By basic algebraic knowledge we can say:
The solutions of equation:
$a{{x}^{2}}+bx+c=0$ are
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
By using this we know
$a=1,b=1,c=1$
By substituting values of a, b, c into expression we get
$\begin{align}
  & x=\dfrac{-1\pm \sqrt{1-4}}{2}=\dfrac{-1\pm \sqrt{-1}\sqrt{3}}{2} \\
 & x=\dfrac{-1\pm i\sqrt{3}}{2} \\
\end{align}$
$\dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2}$ is named as $\omega $
So, $\omega =\dfrac{-1}{2}+\dfrac{i\sqrt{3}}{2}$
And also ${{\omega }^{3}}=1.............(B)$
By substituting $\omega $ into original expression, we get:
${{\omega }^{2}}+\omega +1=0..............(C)$
By substituting $\omega $ into equation (A) we get:
$4+5{{\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}=4+5{{\omega }^{334}}+3{{\omega }^{365}}$
By writing equation in terms of ${{\omega }^{3}}$ , we get:
$4+5{{\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}=4+5{{\left( {{\omega }^{3}} \right)}^{111}}+3{{\left( {{\omega }^{3}} \right)}^{121}}{{\omega }^{2}}$
By substituting equation (B) here, we get:
$4+5{{\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}=4+5\omega +3{{\omega }^{2}}$
By breaking terms, we get:
$4+5{{\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}=3+1+3\omega +2\omega +3{{\omega }^{2}}$
By taking “3” common we get:
$\begin{align}
  & 4+5{{\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}=1+2\omega +3\left( 1+{{\omega }^{2}}+\omega \right) \\
 & =1+2\omega
\end{align}$
By substituting $\omega $ value back, we get:
$4+5{{\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}=1+\left( -1+\sqrt{3}i \right)=\sqrt{3}i$
Therefore, $4+5{{\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{334}}+3{{\left( -\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2} \right)}^{365}}=\sqrt{3}i$
Option (c) is correct.

Note: If in a complex number a + ib, the ratio a : b is $1:\sqrt{3}$ always uses the concept of $\omega $ . Here we can also use the conversion of a given complex number in euler form and then simplifying the Euler form,it would be an easier way to solve this type of question.