MHT CET · Maths · Complex Number
Let \(\left(-2-\frac{1}{3}\right)^3=\frac{x+\mathrm{i} y}{27}, \mathrm{i}=\sqrt{-1}\), where \(x\) and \(y\) are real numbers, then \((y-x)\) has the value
- A -91
- B -85
- C 85
- D 91
Answer & Solution
Correct Answer
(D) 91
Step-by-step Solution
Detailed explanation
\(\left(-2-\frac{1}{3} \mathrm{i}\right)^3=\frac{x+\mathrm{i} y}{27} \)
\( \left(-2-\frac{1}{3} \mathrm{i}\right)^3=\frac{1}{27}(-6-\mathrm{i})^3 \)
\( \text {Consider, }(-6-\mathrm{i})^3 \)
\( =(-6)^3+3(-6)^2(-\mathrm{i})+3(-6)(-\mathrm{i})^2+(-\mathrm{i})^3 \)
\( =-216-108 \mathrm{i}+18+\mathrm{i} \)
\( =-198-107 \mathrm{i} \)
\( \therefore \left(-2-\frac{1}{3} \mathrm{i}\right)^3=\frac{-198-107 \mathrm{i}}{27}\)
Comparing with \(\frac{x+i y}{27}\), we get
\(x=-198, y=-107\)
\(y-x=-107+198=91\)
\( \left(-2-\frac{1}{3} \mathrm{i}\right)^3=\frac{1}{27}(-6-\mathrm{i})^3 \)
\( \text {Consider, }(-6-\mathrm{i})^3 \)
\( =(-6)^3+3(-6)^2(-\mathrm{i})+3(-6)(-\mathrm{i})^2+(-\mathrm{i})^3 \)
\( =-216-108 \mathrm{i}+18+\mathrm{i} \)
\( =-198-107 \mathrm{i} \)
\( \therefore \left(-2-\frac{1}{3} \mathrm{i}\right)^3=\frac{-198-107 \mathrm{i}}{27}\)
Comparing with \(\frac{x+i y}{27}\), we get
\(x=-198, y=-107\)
\(y-x=-107+198=91\)
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