Suppose x and y are two real numbers with x ≠ y, such that
y = 1+x+x
2, and:
x = 1+y+y
2.
Determine the value of x3+y3.
If x=1+y+y^2 and y=1+x+x^2,
Then
x^2=(y-x-1)
y^2=(x-y-1)
Since x^3+y^3 = (x + y) (x^2 - x y + y^2)
x^3+y^3 = (x + y) ((y-x-1) - x y + (x-y-1))
(x + y) (x^2 - x y + y^2) = -(x+y)(xy+2)
So that x^2 + y^2 = -2, with no real solutions for x,y.
So we could stop there, for the problem calls for x,y real.
But continuing,
Then y = +-sqrt(-x^2 - 2)
Substituting
x-sqrt(-x^2 - 2)-1=(-x^2 - 2), x=i
x+sqrt(-x^2 - 2)-1=(-x^2 - 2), x={-i, -1-isqrt(2), -1+isqrt(2)}
Then x^3= {-i, i, 5-sqrt(2)i, 5+sqrt(2)i}
Since x^2=(y-x-1):
{x,y}= {i,i}, {-i,-i}, {-1-isqrt(2),-1+isqrt(2)}, {-1+isqrt(2),-1-isqrt(2)}
But in the first two cases, x=y, which the puzzle prohibits.
Then y^3= {5+sqrt(2)i,5-sqrt(2)i}
and x^3+y^3= {10,10}
Edited on October 18, 2023, 5:08 am
|
|
Posted by broll
on 2023-10-18 05:07:03 |
Possible Solution
