Sure It Was Predictable
⎩ ⎪ ⎨ ⎪ ⎧ x 2 + 4 = y 3 + 4 x − z 3 y 2 + 4 = z 3 + 4 y − x 3 z 2 + 4 = x 3 + 4 z − y 3
Let x , y and z be all real numbers satisfying the system of equations above. Find the value of x ! y ! z !
Notation:
!
is the
factorial
notation. For example,
8
!
=
1
×
2
×
3
×
⋯
×
8
.
The answer is 8.
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2 solutions
Yes same method, but i got 2 question
-
How can we make contradiction for x = y = z
-
Is every x , y , z is always the same regarding to other problem that is similar like mine ?
For completeness, we should verify that ( 2 , 2 , 2 ) is a solution to the original equation.
You have found a necessary condition, but not shown that it is sufficient.
⎩ ⎪ ⎨ ⎪ ⎧ x 2 + 4 = y 3 + 4 x − z 3 y 2 + 4 = z 3 + 4 y − x 3 z 2 + 4 = x 3 + 4 z − y 3
Adding up these three equations, gives us
x ² + y ² + z ² + 1 2 = 4 ( x + y + z ) .
I did some trial and error, and have ( x , y , z ) = ( 2 , 2 , 2 ) .
Hence, our final answer is 8 .
The given equations can be rewritten as
⎩ ⎪ ⎨ ⎪ ⎧ ( x − 2 ) 2 = y 3 − z 3 ( y − 2 ) 2 = z 3 − x 3 ( z − 2 ) 2 = x 3 − y 3
Adding these 3 equations together yields ( x − 2 ) 2 + ( y − 2 ) 2 + ( z − 2 ) 2 = 0 , the sole solution to which is ( x , y , z ) = ( 2 , 2 , 2 ) , (since each of the squares, which are necessarily non-negative, must equal 0 ).
Thus x ! y ! z ! = 8 .
Edit: As Calvin notes, for completeness we need to verify that ( x , y , z ) = ( 2 , 2 , 2 ) satisfies the original equations, which it does indeed.