$(x+y)^{6}$ =64

$(x-y)^{2}$ =16

x+y=?

asked by guest
on Jul 26, 2024 at 8:53 pm



You asked:

Given the following equations: \[\begin{aligned}{\left( x + y \right)}^{6} &= 64\\ {\left( x - y \right)}^{2} &= 16\end{aligned}\] Evaluate \(x + y\).

MathBot Answer:

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

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

By substituting these values into the expression, we find that: $$\begin{aligned}x + y &= -2\end{aligned}$$$$\begin{aligned}x + y &= 2\end{aligned}$$$$\begin{aligned}x + y &= -1 - \sqrt{3} i \approx -1.0 - 1.73205 i\end{aligned}$$$$\begin{aligned}x + y &= -1 + \sqrt{3} i \approx -1.0 + 1.73205 i\end{aligned}$$$$\begin{aligned}x + y &= 1 - \sqrt{3} i \approx 1.0 - 1.73205 i\end{aligned}$$$$\begin{aligned}x + y &= 1 + \sqrt{3} i \approx 1.0 + 1.73205 i\end{aligned}$$

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

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