After multiplying by z, given that we know it is solvable, it must be factorable. Noting that y has degree 1, and that all the coefficients are ±1 makes it fairly easy to find: (x² - x + z)(xz - y + z) = 0
I arrived at the same product, but started by trying to solve for y after multiplying by z. After isolating the y terms, we have y(x² - x + z) = z(x³ + xz - x + z) . It wasn't immediately obvious to me, but with a little fiddling (polynomial division or adding & subtracting x²), we can show that x³ + xz - x + z = (x+1)(x² - x + z). We don't want to divide through by the common polynomial and lose solutions, but moving everything to the LHS and factoring gives the product.
I was able to compute the discriminate: sqrt(x^6 - 2x^4 + 2yx^3 - x^2 + 2yx + y^2) By taking the square root of a polynomial directly and got: x^3 - x + y Hooray! We got the same answer. However, as a word of caution, this may not have been the case since (x^3 + bx + y) in your solution, may have ended up being (x^3 + bx - y) as the actual final solution.
![Seungmin "그렇게, 천천히, 우리(As we are)" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/kAzmhLHePqU/mqdefault.jpg)
Is nice to see new colors on the thumbnail, also love the question but more the solution!
After multiplying by z, given that we know it is solvable, it must be factorable. Noting that y has degree 1, and that all the coefficients are ±1 makes it fairly easy to find: (x² - x + z)(xz - y + z) = 0
I arrived at the same product, but started by trying to solve for y after multiplying by z.
After isolating the y terms, we have y(x² - x + z) = z(x³ + xz - x + z) .
It wasn't immediately obvious to me, but with a little fiddling (polynomial division or adding & subtracting x²),
we can show that x³ + xz - x + z = (x+1)(x² - x + z).
We don't want to divide through by the common polynomial and lose solutions, but moving everything to the LHS and factoring gives the product.
A beautiful equation indeed, and so is your solution!
Thank you! 🧡
amazing equation and beautiful and very clean solution, very well explained and easy to follow
Glad you liked it! 🧡
I was able to compute the discriminate:
sqrt(x^6 - 2x^4 + 2yx^3 - x^2 + 2yx + y^2)
By taking the square root of a polynomial directly and got:
x^3 - x + y
Hooray! We got the same answer. However, as a word of caution, this may not have been the case since (x^3 + bx + y) in your solution, may have ended up being (x^3 + bx - y) as the actual final solution.
Really nice question 😊😊😊😊😊❤❤❤
Thanks 😊🧡
Transforming the initial equation leads to a nice factorization:
[x(x-1)+z][z(x+1)-y]=0
=>
a) x(x-1)+z=0
b) z(x+1)-y=0
😉
It may be easy to go from y^2 term
y^2-2y(x^3+x)+(x^3+x)^2
Why is this YT channel named “Shorts”? 😂
Because I upload shorts time to time 😜😁
@@ShortsOfSybershort and long are relative rather than absolute