Solve for x:
3√x - 3√y = 10
xy = {(8-x-y)/6}3
let a = ∛x and b = ∛y
a - b = 10
(6ab)^3 = (8 - a^3 - b^3)^3
6ab = 8 - a^3 - b^3
a^3 + b^3 + 6ab - 8 = 0
substitute b = a-10
a^3 + (a-10)^3 + 6a(a-10) - 8 = 0
2a^3 - 30a^2 + 300a - 1000 + 6a^2 - 60a - 8 = 0
2a^3 - 24a^2 + 240a - 1008 = 0
a^3 - 12a^2 + 120a - 504 = 0
f(a) = a^3 - 12a^2 + 120a - 504
The factors of 504 are ± [1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 18, 21, 24, 28, 36, 42, 56, 63, 72, 84, 126, 168, 252, 504]. Testing all 48 possibilities shows that only +6 is a zero of the function of a.
a=6
Since a-b=10, b=-4
But a and b are the cube roots of x and y.
So (x,y) = (216, -64)
This solves both original equations
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Posted by Larry
on 2024-10-24 20:26:05 |
Solution