Case I : x=y=z then x²+x=2 then x=1,-2 and solution x=y=z={1,-2} Case II x≠y≠z But how we prove this?? The best polite proffesor in youtube my greeting
x + yz = 2 #1 y + zx = 2 #2 z + xy = 2 #3 #1 is equivalent to: x^2 + xyz = 2x (x - 1)^2 = 1 - xyz Similarly, we have: (x - 1)^2 = (y - 1)^2 = (z - 1)^2 = 1 - xyz #4 #1 - #3: x + yz - z - xy = 0 (x - z)(1 - y) = 0 Similarly, we have: (y - x)(1 - z) = (z - y)(1 - x) = 0 x = y = z or x = y = z = 1 Consider when x = y = z =/= 1: We can rewrite #1 as: x^2 + x - 2 = 0 (x + 2)(x - 1) = 0 x = y = z = -2 Hence {x, y, z} = {1, 1, 1} or {-2, -2, -2} QED [ Fun algebraic question :) ]
Very nice explanation
Thank you so much my friend for your support👍👍👍
Thank you sir
Thanks a lot my friend for your continued support! I really appreciate it👍👍👍
Very nice explanation prof.
Thank you my friend haha👍👍👍
Another great video prof🎉
Thansk a lot my friend for your support👍👍👍
I dont like synthetic division but long division just like you prof. Good thing I do not see uneducated comment now
Haha thanks for your support my friend👍👍👍
Case I : x=y=z then
x²+x=2 then x=1,-2
and solution x=y=z={1,-2}
Case II
x≠y≠z
But how we prove this??
The best polite proffesor in youtube my greeting
Hello my friend. I used further factoring for case 2. I always appreciate your support👍👍👍
Fabulous explanation - Perhaps missing a statement re symmetry or is this picked up by x = y = z?
I do watch to the end - but the new video choices do block left of board.
Hello my friend! Yes symmetry was the key in my solution development haha thanks for your comment👍👍👍
x + yz = 2 #1
y + zx = 2 #2
z + xy = 2 #3
#1 is equivalent to:
x^2 + xyz = 2x
(x - 1)^2 = 1 - xyz
Similarly, we have:
(x - 1)^2 = (y - 1)^2 = (z - 1)^2 = 1 - xyz #4
#1 - #3:
x + yz - z - xy = 0
(x - z)(1 - y) = 0
Similarly, we have:
(y - x)(1 - z) = (z - y)(1 - x) = 0
x = y = z or x = y = z = 1
Consider when x = y = z =/= 1:
We can rewrite #1 as:
x^2 + x - 2 = 0
(x + 2)(x - 1) = 0
x = y = z = -2
Hence {x, y, z} = {1, 1, 1} or {-2, -2, -2}
QED [ Fun algebraic question :) ]
Nice solution my friend! haha thanks for sharing it👍👍👍