(x+y)^6 = 64

(x-y)^2 = 16

asked by guest
on Nov 17, 2024 at 7:54 am



You asked:

Solve the system of equations:\[\begin{aligned}{\left( x + y \right)}^{6} &= 64\\{\left( x - y \right)}^{2} &= 16\end{aligned}\]

MathBot Answer:

The system of equations has \(12\) solutions.

\[x = -3, y = 1\]\[x = -1, y = 3\]\[x = 1, y = -3\]\[x = 3, y = -1\]\[x = - \frac{5}{2} - \frac{\sqrt{3} i}{2}, y = \frac{3}{2} - \frac{\sqrt{3} i}{2}\]\[x = - \frac{5}{2} + \frac{\sqrt{3} i}{2}, y = \frac{3}{2} + \frac{\sqrt{3} i}{2}\]\[x = - \frac{3}{2} - \frac{\sqrt{3} i}{2}, y = \frac{5}{2} - \frac{\sqrt{3} i}{2}\]\[x = - \frac{3}{2} + \frac{\sqrt{3} i}{2}, y = \frac{5}{2} + \frac{\sqrt{3} i}{2}\]\[x = \frac{3}{2} - \frac{\sqrt{3} i}{2}, y = - \frac{5}{2} - \frac{\sqrt{3} i}{2}\]\[x = \frac{3}{2} + \frac{\sqrt{3} i}{2}, y = - \frac{5}{2} + \frac{\sqrt{3} i}{2}\]\[x = \frac{5}{2} - \frac{\sqrt{3} i}{2}, y = - \frac{3}{2} - \frac{\sqrt{3} i}{2}\]\[x = \frac{5}{2} + \frac{\sqrt{3} i}{2}, y = - \frac{3}{2} + \frac{\sqrt{3} i}{2}\]

\(i\) is the imaginary unit, defined as \(i^2 = -1\).

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