Animated Solution for Mathematics - Complex Numbers: Let $(\frac{-2}{3}+\frac{1}{3}i{)}^3=\frac{x+iy}{27}(i=\sqrt{-1})$, where $x$ and $y$ are real numbers, then $y-x$ equals :

Imagine you are standing on the precipice of a complex algebraic expression. You see ${\left(-\frac{2}{3}+\frac{1}{3}i\right)}^3$ and your first instinct might be to dive straight into the expansion.

But wait! As an elite JEE aspirant, you know that the path of least resistance is often the path of greatest insight. We are given the cube of a complex number, and we need to compare it with a standard form to find the values of $x$ and $y$.

To make the calculation easier, let us take $-\frac{1}{3}$ common from inside the bracket. Expanding a cube with fractions is difficult and prone to errors.

By pulling out the constant, we transform the expression into:

Now, when we apply the cube to both terms, we get:

Notice how the denominator $27$ appears naturally? This is the elegance of math—the structure of the problem guides you toward the solution.

To expand $(2-i{)}^3$, we use the standard identity $(a-b{)}^3=a^3-3a^{2}b+3a{b}^2-b^3$. Here, $a=2$ and $b=i$.

Let us break it down:

1. $a^3=2^3=8$ 2. $3a^{2}b=3(2{)}^2(i)=12i$ 3. $3a{b}^2=3(2)(i{)}^2=6(-1)=-6$ 4. $b^3=i^3=-i$

Combining these terms, we get:

We are almost there! We substitute this back into our factored expression:

Now, we compare this with the given form $\frac{x+iy}{27}$. By equating the real and imaginary parts, we find:

The final step is to calculate $y-x$. Substituting our values, we get:

And there it is! The complexity dissolves into a clean, satisfying integer. Remember, in the heat of the exam, stay calm, look for the common factors, and trust your algebraic foundations.