Let $\left(-2 - \dfrac{1}{3}i\right)^3 = \dfrac{x + iy}{27}$ $(i = \sqrt{-1})$, where $x$ and $y$ are real numbers, then $y - x$ equals
Step-by-Step Solution
Key Concept: The binomial cube formula avoids the tedious step-by-step multiplication; keep track of the $i^2 = -1$ sign at each power.
**Step 1: Expand using Binomial theorem**
Using $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ with $a = -2$, $b = -\frac{i}{3}$:
$a^3 = -8$, $3a^2b = 3(4)\!\left(-\frac{i}{3}\right) = -4i$, $3ab^2 = 3(-2)\!\left(-\frac{1}{9}\right) = \frac{2}{3}$, $b^3 = \frac{i}{27}$.
**Step 2: Collect real and imaginary parts**
Sum $= \left(-8 + \dfrac{2}{3}\right) + i\!\left(-4 + \dfrac{1}{27}\right) = -\dfrac{22}{3} - \dfrac{107}{27}i$.
**Step 3: Read off x, y and compute y − x**
$\dfrac{x+iy}{27} = -\dfrac{22}{3} - \dfrac{107i}{27}$, so $x = -198$, $y = -107$. Thus $y - x = -107 - (-198) = 91$.
Correct Answer: 1