Question

Let ${{\left( -2-\dfrac{1}{3}i \right)}^{3}}=\dfrac{x+iy}{27}\text{ }\left( i=\sqrt{-1} \right)$, where x and y are real numbers, then $y-x$ equals:
$\begin{align}
  & \text{A}\text{. -85} \\
 & \text{B}\text{. 85} \\
 & \text{C}\text{. -91} \\
 & \text{D}\text{. 91} \\
\end{align}$

Answer 1

Hint: We have given an expression ${{\left( -2-\dfrac{1}{3}i \right)}^{3}}=\dfrac{x+iy}{27}\text{ }\left( i=\sqrt{-1} \right)$. Now, first we simplify L.H.S. of the given equation ${{\left( -2-\dfrac{1}{3}i \right)}^{3}}$ by using the formula ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}$. Then, compare the L.H.S. and R.H.S. of the given equation to get the values of $x$ and $y$. Then, put the values in $y-x$ and solve to find the value.

Complete step-by-step answer:
We have been given an equation ${{\left( -2-\dfrac{1}{3}i \right)}^{3}}=\dfrac{x+iy}{27}\text{ }\left( i=\sqrt{-1} \right)$, where x and y are real numbers.
We have to find the value of $y-x$.
Now, let us first take the L.H.S. of the given equation.
$\Rightarrow {{\left( -2-\dfrac{1}{3}i \right)}^{3}}$or $\Rightarrow -{{\left( 2+\dfrac{1}{3}i \right)}^{3}}$
Now, we know that ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}$
Here, we have $a=2,b=\dfrac{i}{3}$
Now, substituting the values, we get
\[\Rightarrow -{{\left( 2+\dfrac{i}{3} \right)}^{3}}=-\left( {{2}^{3}}+{{\left( \dfrac{i}{3} \right)}^{3}}+3\times {{2}^{2}}\times \dfrac{i}{3}+3\times 2\times {{\left( \dfrac{i}{3} \right)}^{2}} \right)\]
Now, simplifying the above equation we get
\[\begin{align}
  & \Rightarrow -{{\left( 2+\dfrac{i}{3} \right)}^{3}}=-\left( 8+{{\left( \dfrac{i}{3} \right)}^{3}}+3\times 4\times \dfrac{i}{3}+3\times 2\times {{\left( \dfrac{i}{3} \right)}^{2}} \right) \\
 & \Rightarrow -{{\left( 2+\dfrac{i}{3} \right)}^{3}}=-\left( 8+\dfrac{{{i}^{3}}}{{{3}^{3}}}+4i+3\times 2\times \dfrac{{{i}^{2}}}{{{3}^{2}}} \right) \\
 & \Rightarrow -{{\left( 2+\dfrac{i}{3} \right)}^{3}}=-\left( 8+\dfrac{{{i}^{3}}}{27}+4i+2\times \dfrac{{{i}^{2}}}{3} \right) \\
\end{align}\]
Now, take L.C.M. to solve further, we get
\[\begin{align}
  & \Rightarrow -{{\left( 2+\dfrac{i}{3} \right)}^{3}}=-\left( \dfrac{8\times 27+{{i}^{3}}+27\times 4i+9\times 2{{i}^{2}}}{27} \right) \\
 & \Rightarrow -{{\left( 2+\dfrac{i}{3} \right)}^{3}}=-\left( \dfrac{216+{{i}^{3}}+108i+18{{i}^{2}}}{27} \right) \\
\end{align}\]
Now, we know that the values of imaginary number will be
$\begin{align}
  & {{i}^{2}}=-1 \\
 & {{i}^{3}}=-i \\
\end{align}$
When we substitute values, we get
\[\Rightarrow -{{\left( 2+\dfrac{i}{3} \right)}^{3}}=-\left( \dfrac{216-i+108i-18}{27} \right)\]
When we simplify the above equation, we get
\[\Rightarrow -{{\left( 2+\dfrac{i}{3} \right)}^{3}}=-\left( \dfrac{198+107i}{27} \right)\]
Now, substitute the value in the given equation we have
$\begin{align}
  & {{\left( -2-\dfrac{1}{3}i \right)}^{3}}=\dfrac{x+iy}{27} \\
 & \Rightarrow -\left( \dfrac{198+107i}{27} \right)=\dfrac{x+iy}{27} \\
 & \Rightarrow \left( \dfrac{-198-107i}{27} \right)=\dfrac{x+iy}{27} \\
\end{align}$
When we compare L.H.S. and R.H.S. of the given equation, we get the values
$\begin{align}
  & x=-198 \\
 & y=-107 \\
\end{align}$
Now, we have to calculate the value of $y-x$ as asked in the question, we get
$\begin{align}
  & y-x=-107-\left( -198 \right) \\
 & y-x=-107+198 \\
 & y-x=91 \\
\end{align}$
So, the value of $y-x$ is $91$.

So, the correct answer is “Option D”.

Note: A complex number is a combination of real number and imaginary number. In such types of questions we need the values of imaginary numbers, so always keep in mind the common values of imaginary numbers to solve easily. As the calculation is complex and lengthy so be careful while solving. The identity used is ${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}$ so please be careful to write the expansion without making change in signs of the terms of the expansion.