এবং    হলে , x²+xy+y² এর মান-

দেওয়া আছে,

\(x = \frac{1}{2}(-1 + \sqrt{-3})\)

\(y = \frac{1}{2}(-1 - \sqrt{-3})\)

আমরা জানি, \(\sqrt{-3} = \sqrt{3}i\)

সুতরাং,

\(x = \frac{1}{2}(-1 + \sqrt{3}i)\)

\(y = \frac{1}{2}(-1 - \sqrt{3}i)\)

এখন, \(x + y = \frac{1}{2}(-1 + \sqrt{3}i) + \frac{1}{2}(-1 - \sqrt{3}i) = \frac{1}{2}(-1 + \sqrt{3}i - 1 - \sqrt{3}i) = \frac{1}{2}(-2) = -1\)

এবং, \(xy = \frac{1}{2}(-1 + \sqrt{3}i) \cdot \frac{1}{2}(-1 - \sqrt{3}i) = \frac{1}{4}[(-1)^2 - (\sqrt{3}i)^2] = \frac{1}{4}[1 - 3i^2] = \frac{1}{4}[1 - 3(-1)] = \frac{1}{4}[1 + 3] = \frac{1}{4} \cdot 4 = 1\)

এখন, \(x^2 + xy + y^2 = (x + y)^2 - xy = (-1)^2 - 1 = 1 - 1 = 0\)

অতএব, \(x^2 + xy + y^2\) এর মান 0। 🎉