+87 Let \(x\) and \(y\) be real numbers whose absolute values are different and that satisfy \( \begin{align*} x^3 &= 20x + 7y \\ y^3 &= 7x + 20y. \end{align*}\)
Find \(xy.\)
I used a graphical approach (desmos) to graph the two equations you will see the following sets of coordinates satisfy the equations :
x,y = (-1.7 , 4.1) (-4.1, 1.7) and (4.1, -1.7) and (1.7, - 4.1)
0,0 -3.6, 3.6 and 3.6, -3.6 5.2 , 5.2 -5.2, -5.2 also work, but the absolute values of x and y are equal at these points, so they do not satisfy the question.
Here is the graph:
Let xy=k, so that y = k/x.
Substitute this into the first equation to get \(x^2 - 20x^2 - 7k = 0\)
Solve this (using the usual formula) as a quadratic in x squared to get
\(x^2 = 10 \pm \sqrt{100 + 7k}\)
The original equations are symmetric in x and y so repeating the procedure will produce an identical result for y.
Since x and y are different, one will have the positive sign in the middle, the other the negative sign.
Multiplying the two results produces (difference between two squares)
\(x^2 y^2 = 100 - (100 + 7k) = -7xy\)
Since xy is not equal to 0, it follows that xy = -7.
You can solve for y in the first equation, to get y = (x^3 - 20x)/7.
You can then substitute into the second equations, which gives you
\(\left( \frac{x^3 - 20x}{7} \right)^3 = 7x + 20 \left( \frac{x^3 - 20x}{7} \right)\)
This factors as x(x - 5)(x + 5)(x^2 + 12)(x^4 - 20x^2 + 49) = 0. You can then work out the rest.
