Let x and y be real numbers whose absolute values are different and that satisfy

x^3 = 20x + 7y

y^3 = 7x + 20y

Find xy.

Thanks so much!

AnonymousConfusedGuy Jan 21, 2018

#1**+1 **

I will assume you want to find x,y where x and y have different absolute values...(NOT x*y as written in your question)

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:

ElectricPavlov Jan 21, 2018

#3**+1 **

Thanks! I actually was looking for x*y, however this was helpful nonetheless!

AnonymousConfusedGuy
Jan 21, 2018

#2**+2 **

I believe this problem has appeared several times before....but, anyway

x^3 = 20x + 7y

y^3 = 7x + 20y first, add these

x^3 + y^3 = 27x + 27y

(x + y)( x^2 - xy + y^2) = 27(x + y) divide by x + y

x^2 - xy + y^2 = 27 (1)

Next, subtract the first two equations

x^3 - y^3 = 13y - 13y

(x - y) (x^2 + xy+ y^2) = 13(x - y) divide by x - y

x^2 + xy + y^2 = 13 (2)

Subtract (2) from (1)

-2xy = 14 divide both sides by -2

xy = -7

CPhill Jan 21, 2018

#5**0 **

No problem ACG....I wouldn't expect anyone to scroll through the thousands of questions on here to see if theirs has been asked before.....I just happened to remember this one....to give full credit, I believe that heureka or alan may have solved it originally....

CPhill Jan 21, 2018

#6**+1 **

Oh...Ok sorry.... If you multiply all of my x.y 's together (rounded) they come out to -7

ElectricPavlov Jan 21, 2018