Question

Get x, y from below equation and find value of y – x, where x and y are real numbers
\[{{\left( -2-\dfrac{1}{3}i \right)}^{3}}=\dfrac{x+iy}{27},\left( i=\sqrt{-1} \right)\]
(a)-85
(b)85
(c)-91
(d)91

Answer 1

Hint: First solve the left hand side like a normal algebraic equation. Then multiply and divide by 27 then compare with the right hand side to find values of x, y.
You can use distributive property:
b.(a + c) = b.a + b.c

Complete step-by-step solution -

First we need to separate out the left hand side.
\[L.H.S.={{\left( -2-\dfrac{1}{3}i \right)}^{3}}\]
Now write it as a multiplication of 3 same terms.
\[{{\left( -2-\dfrac{1}{3}i \right)}^{3}}=\left( -2-\dfrac{1}{3}i \right).\left( -2-\dfrac{1}{3}i \right).\left( -2-\dfrac{1}{3}i \right)\]
By treating 2 multiplied terms as one entity we can apply distributive law:
b.(a + c) = b.a + b.c

By applying above law, we get:
\[=-2.\left( -2-\dfrac{i}{3} \right).\left( -2-\dfrac{i}{3} \right)-\dfrac{i}{3}.\left( -2-\dfrac{i}{3} \right).\left( -2-\dfrac{i}{3} \right)\]
Now you can take one term as common and apply distributive law again.
\[=\left( -2-\dfrac{i}{3} \right)\left( -2.\left( -2-\dfrac{i}{3} \right)-\dfrac{i}{3}.\left( -2-\dfrac{i}{3} \right) \right)\]
By applying distributive law twice inside the bracket we get:
\[=\left( -2-\dfrac{i}{3} \right)\left( \left( -2.-2 \right)+\left( -2.\dfrac{-i}{3} \right)+\left( -2.\dfrac{-i}{3} \right)+\left( \dfrac{-i}{3}.\dfrac{-i}{3} \right) \right)\]
By simplifying, we get:
\[\begin{align}
  & =\left( -2-\dfrac{i}{3} \right)\left( 4+\dfrac{2i}{3}+\dfrac{2i}{3}+\dfrac{{{i}^{2}}}{9} \right) \\
 & =\left( -2-\dfrac{i}{3} \right)\left( 4+\dfrac{4i}{3}+\dfrac{{{i}^{2}}}{9} \right) \\
\end{align}\]
We know:
i is solution of equation:
\[{{i}^{2}}=-1\]
By substituting above equation and simplifying, we get:
\[\begin{align}
  & =\left( -2-\dfrac{i}{3} \right)\left( 4+\dfrac{4i}{3}-\dfrac{1}{9} \right) \\
 & =\left( -2-\dfrac{i}{3} \right)\left( \dfrac{35}{9}+\dfrac{4i}{3} \right) \\
\end{align}\]
By treating one of the both terms as one entity apply distributive law.
b.(a + c) = b.a + b.c
\[=\left( -2 \right)\left( \dfrac{35}{9}+\dfrac{4i}{3} \right)+\left( -\dfrac{i}{3} \right)\left( \dfrac{35}{9}+\dfrac{4i}{3} \right)\]
By applying distributive law twice with each term, we get:
\[=\left( -2.\dfrac{35}{9} \right)+\left( -2.\dfrac{4i}{3} \right)+\left( \dfrac{-i}{3}.\dfrac{35}{9} \right)+\left( \dfrac{-i}{3}.\dfrac{4i}{3} \right)\]
We know,
i is solution of equation:
\[{{i}^{2}}=-1\]
By substituting above equation and simplifying, we get:
\[=\left( -\dfrac{70}{9} \right)+\left( \dfrac{-8i}{3} \right)+\left( \dfrac{-35i}{27} \right)+\left( \dfrac{4}{9} \right)\]
Now adding the real terms and imaginary terms separately, we get:
\[\begin{align}
  & =\left( -\dfrac{66}{9} \right)+\left( \dfrac{-107i}{27} \right) \\
 & =\left( -\dfrac{22}{3} \right)+\left( \dfrac{-107i}{27} \right) \\
\end{align}\]
Finally,
\[\text{left hand side = }\left( -\dfrac{22}{3} \right)+\left( \dfrac{-107i}{27} \right)\text{ }\]
Now taking right hand side into consideration, we get:
\[\text{right hand side = }\dfrac{x+iy}{27}\]
By separating real and imaginary terms, we get:
\[\text{right hand side = }\dfrac{x}{27}+\dfrac{iy}{27}\]
By equating left hand side and right hand side, we get:
\[\left( -\dfrac{22}{3} \right)+\left( \dfrac{-107i}{27} \right)\text{=}\dfrac{x}{27}\text{+}\dfrac{iy}{27}\text{ }\]
By equating real terms, we get:
\[\dfrac{x}{27}=-\dfrac{22}{3}\]
By multiplying 27 on both sides, we get:
\[\begin{align}
  & \dfrac{x}{27}\times 27=\dfrac{-22}{3}\times 27 \\
 & \\
\end{align}\]
By equating imaginary terms, we get:
\[\dfrac{iy}{27}=-\dfrac{107i}{27}\]
By multiplying 27 on both sides and cancelling i, we get:
\[\begin{align}
  & \dfrac{iy}{27}\times 27=\dfrac{-107i}{27}\times 27 \\
 & \\
\end{align}\]
 We need y – x.
By above values we can say:
y – x = -107 – (-198)
= -107 + 198
= 91
\[\therefore \]The value of y – x is 91.
Option (d) is correct.

Note: Don’t take the left hand side directly, first you have to multiply 27 or else you’ll get a different answer which is also present in options. This way you may lead to the wrong answer.
Alternative method is to apply a algebraic identity:
\[{{\left( a+b \right)}^{3}}={{a}^{3}}+3ab\left( a+b \right)+{{b}^{3}}\].