Evaluate the following, expressing your answer in cartesian form $\left( a+bi \right):{{\left( 1-3i \right)}^{3}}$.

Hint: We know that the cartesian form of complex numbers is nothing but the way of representing a complex number in the form of $\left( a+bi \right)$, where $a,b$ are real numbers. We also know that, we can write, ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$. So, we have to use this in the question to get the cartesian form of the complex number.

Complete step-by-step answer:
In this question, we have been asked to find the cartesian form of complex number ${{\left( 1-3i \right)}^{3}}$, that is we have to express ${{\left( 1-3i \right)}^{3}}$ in the form of $\left( a+bi \right)$. Now, we know that ${{\left( a-b \right)}^{3}}$ can be written as, ${{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$, or as ${{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}$. Now we will use this formula to simplify ${{\left( 1-3i \right)}^{3}}$. By simplifying, we get,
$\begin{align}
  & {{\left( 1-3i \right)}^{3}}={{\left( 1 \right)}^{3}}-{{\left( 3i \right)}^{3}}-3{{\left( 1 \right)}^{2}}\left( 3i \right)+3\left( 1 \right){{\left( 3i \right)}^{2}} \\
 & \Rightarrow {{\left( 1-3i \right)}^{3}}=1-27{{i}^{3}}-9i+27{{i}^{2}} \\
\end{align}$
Now, we know that ${{i}^{2}}=-1$ and that, ${{i}^{3}}=-i$. By substituting them in the above equation, we get,
$\begin{align}
  & {{\left( 1-3i \right)}^{3}}=1-27\left( -i \right)-9i+27\left( -1 \right) \\
 & \Rightarrow {{\left( 1-3i \right)}^{3}}=1+27i-9i-27 \\
\end{align}$
By adding all the like terms in the above equation, we get,
$\begin{align}
  & {{\left( 1-3i \right)}^{3}}=\left( 1-27 \right)+\left( 27-9 \right)i \\
 & \Rightarrow {{\left( 1-3i \right)}^{3}}=-26+18i \\
\end{align}$
After simplification, we get the value of ${{\left( 1-3i \right)}^{3}}$as $-26+18i$. Hence the expression ${{\left( 1-3i \right)}^{3}}$ can be written in the cartesian form as $-26+18i$, where $\left( -26 \right)$ is the real part of the cartesian form and $\left( 18i \right)$ is the imaginary part of the cartesian form.

Note: The possible mistakes the students can do while solving this type of questions are by getting confused with the terms like, ${{i}^{2}}$ or terms like ${{i}^{3}}$. The students can also make a mistake of writing, ${{i}^{2}}=1$ and ${{i}^{3}}=i$ while trying to solve the question in a hurry, which will give them a wrong answer.