Answer
$(-5, -8)$
$(8, 5)$
Work Step by Step
Given the system of equations:
1. $x - y = 3$
2. $x^3 - y^3 = 387$
Reformat equation 2
1. $x - y = 3$
2. $(x-y) (x^2 + xy + y^2) = 387$
The question asks to find all real solutions.
Substitute $x - y = 3$
$3(x^2 + xy + y^2) = 387$
$x^2 + xy + y^2 = 129$
$x^2 -2xy + y^2 = 129 - 3xy$
$(x-y)^2 = 129 - 3xy$
$3^2 = 129 - 3xy$
$40 = xy$
Since $x = 3+y$ from equation 1, substitute!
$40 = (3+y)(y)$
$40 = 3y + y^2$
$y^2 + 3y - 40 = 0$
$(y + 8)(y - 5) = 0$
$y = -8, 5$
Substitute the values into the equation
$x = 3 + (-8)$
$x = -5$
$x = 3 + 5$
$x = 8$
All Solutions:
$(-5, -8)$
$(8, 5)$
