AP EAMCET · Maths · Complex Number
If \(1, \omega, \omega^2\) are the cube roots of unity then the value of \((x+y)^2+\left(x \omega+y \omega^2\right)^2+\left(x \omega^2+y \omega\right)^2\) is
- A \(2 x^2 \cdot 3 y^2\)
- B \(4 x y\)
- C \(6 x y\)
- D \(2 x^2 \cdot 2 y^2\)
Answer & Solution
Correct Answer
(C) \(6 x y\)
Step-by-step Solution
Detailed explanation
Given that \(1, \mathrm{w}, \mathrm{w}^2\) are the cube roots of unity. \(\therefore 1+\mathrm{w}+\mathrm{w}^2=0\) and \(\mathrm{w}^3=1\). Now, \((x+y)^2\left(x w+y w^2\right)+\left(x w^2+y w\right)^2\)…
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