Answer:
(3 - √6, 3 + √6) and (3 + √6, 3 - √6).
Step-by-step explanation:
To find an ordered pair (x, y) that satisfies the given equations, we can use the fact that:
x^3 + y^3 = (x + y)(x^2 - xy + y^2).
Substituting the given values, we have:
162 = 6(x^2 - xy + y^2).
Dividing both sides by 6, we get:
27 = x^2 - xy + y^2.
Now we have a system of equations:
x + y = 6,
x^2 - xy + y^2 = 27.
We can solve this system by substitution or elimination.
Let's solve it using the substitution method:
From the first equation, we can express x in terms of y:
x = 6 - y.
Substituting this expression for x in the second equation, we have:
(6 - y)^2 - (6 - y)y + y^2 = 27.
Simplifying the equation:
36 - 12y + y^2 - 6y + y^2 + y^2 = 27,
3y^2 - 18y + 9 = 0.
Dividing both sides by 3, we get:
y^2 - 6y + 3 = 0.
This quadratic equation can be solved using the quadratic formula:
y = [6 ± √(6^2 - 4(1)(3))] / (2(1)).
Simplifying further:
y = [6 ± √(36 - 12)] / 2,
y = [6 ± √24] / 2,
y = [6 ± 2√6] / 2,
y = 3 ± √6.
Now we can find the corresponding values of x using the equation x = 6 - y:
For y = 3 + √6:
x = 6 - (3 + √6) = 3 - √6.
For y = 3 - √6:
x = 6 - (3 - √6) = 3 + √6.
Therefore, the ordered pairs (x, y) that satisfy the given equations are:
(3 - √6, 3 + √6) and (3 + √6, 3 - √6).
