Simplify the expression: $\dfrac{1}{{{i}^{3}}}$.

Hint: Observe that $i$ is a square root of unity. Simplify the expression of the form $\dfrac{1}{x+iy}$ by multiplying and dividing it by $x-iy$. Calculate the value of the expression using the algebraic identity $\left( x+y \right)\left( x-y \right)={{x}^{2}}-{{y}^{2}}$. Further use the fact that as $i=\sqrt{-1}$, we have ${{i}^{4}}=1$.

Complete step-by-step solution -
We have to calculate the value of $\dfrac{1}{{{i}^{3}}}$. We observe that this is a complex number.
We also know that $i$ is a square root of unity. Thus, we have $i=\sqrt{-1}$.
We will first calculate the value of ${{i}^{3}}$.
As we know that $i=\sqrt{-1}$. Thus, we have ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1$.
So, we have ${{i}^{3}}={{i}^{2+1}}={{i}^{2}}\times i=-1\times i=-i$.
Thus, we can rewrite the expression $\dfrac{1}{{{i}^{3}}}$ as $\dfrac{1}{{{i}^{3}}}=\dfrac{1}{-i}$.
We will now further simplify this expression.
We know that we can simplify the expression of the form $\dfrac{1}{x+iy}$ by multiplying and dividing it by $x-iy$.
Substituting $x=0,y=-1$ in the above expression, we can rewrite $\dfrac{1}{{{i}^{3}}}=\dfrac{1}{-i}$ as $\dfrac{1}{{{i}^{3}}}=\dfrac{1}{-i}=\dfrac{i}{-i\times i}$.
We know that ${{i}^{2}}={{\left( \sqrt{-1} \right)}^{2}}=-1$. Substituting this value in the above expression, we have $\dfrac{1}{{{i}^{3}}}=\dfrac{1}{-i}=\dfrac{i}{-i\times i}=\dfrac{i}{-{{i}^{2}}}=\dfrac{i}{-\left( -1 \right)}$.
Thus, we have $\dfrac{1}{{{i}^{3}}}=\dfrac{1}{-i}=\dfrac{i}{-i\times i}=\dfrac{i}{-{{i}^{2}}}=\dfrac{i}{-\left( -1 \right)}=\dfrac{i}{1}=i$.
Hence, the value of the expression $\dfrac{1}{{{i}^{3}}}$ is $i$.

Note: We can also solve this question by multiplying and dividing the expression $\dfrac{1}{{{i}^{3}}}$ by $i$ and then use the fact that as $i=\sqrt{-1}$, we have ${{i}^{4}}={{\left( {{i}^{2}} \right)}^{2}}={{\left( -1 \right)}^{2}}=1$ to calculate the value of the given expression. We can write any complex number in the form $a+ib$, where $ib$ is the imaginary part and $a$ is the real part. We can’t solve this question without using algebraic identities. We must simplify the complex part in the denominator of a fraction by rearranging the terms.