Question

If $x,y \in \mathbb{R}$ and $x+iy=-(6+i)^{3}, i^{2}=-1$, then $x-y$ is equal to ________.

• A. 93
• B. -93
• C. 91
• D. -91
• E. -107

Hint:

$(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
We need to find the values of real numbers \(x\) and \(y\) by first expanding the complex number \((6+i)^3\), applying the negative sign, and then equating the real and imaginary parts of the resulting complex number with \(x + iy\). Finally, we calculate the value of \(x - y\).
Step 2: Key Formula or Approach
We use the binomial expansion formula for a cube: \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\). We also use the properties of the imaginary unit: \(i^2 = -1\) and \(i^3 = i^2 \cdot i = -i\).
Step 3: Detailed Explanation
1. Expand \((6+i)^3\).
Using the formula with \(a=6\) and \(b=i\):
\[ (6+i)^3 = (6)^3 + 3(6)^2(i) + 3(6)(i)^2 + (i)^3 \] \[ = 216 + 3(36)(i) + 18(-1) + (-i) \] \[ = 216 + 108i - 18 - i \] 2. Combine the real and imaginary parts.
\[ (6+i)^3 = (216 - 18) + (108 - 1)i \] \[ = 198 + 107i \] 3. Apply the negative sign.
We are given \(x + iy = -(6+i)^3\).
\[ x + iy = -(198 + 107i) \] \[ x + iy = -198 - 107i \] 4. Equate the real and imaginary parts to find x and y.
By comparing the left and right sides of the equation:
The real part: \(x = -198\)
The imaginary part: \(y = -107\)
5. Calculate \(x - y\).
\[ x - y = (-198) - (-107) \] \[ x - y = -198 + 107 \] \[ x - y = -91 \] Step 4: Final Answer
The value of \(x - y\) is -91.