Solving a Cubic Polynomial System in Two Ways

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Knowing that we immediately arrive at (x+y)(y+z)(x+z) = 0 from where we get x=-y or y=-z or x=-z using which on the original 2 equations trivially yields the result.

For those wondering about the first equality - it's an identity. The way I think best to see why it's the case is:
P = (x+y+z)³-(x³+y³+z³) = (x+y+z)³ - x³ - (y³+z³) = (x+y+z-x)(some stuff) - (y+z)(some stuff) = (y+z)(some stuff) - (y+z)(some stuff) = (y+z)(some stuff) which shows that (y+z) divides the original polynomial P. But we chose to separate x³ arbitrary, by symmetry the same applies to y or z, so (x+z) and (x+y) are divisors too. However P is a degree 3 polynomial thus it can have at most 3 linear factors and we've just found all of them. Hence P = C(x+y)(y+z)(x+z) and to find that C=3 just use x=y=z=1.

randomjin

Nice. Can be easily solved if u know high school algebra.

cube

Wait what...i was expecting an April Fool's Joke...unless I was the joke all along 🤔

mindstreamx

I love that this used Vieta's formulae. I thought I was so clever for using them once I'd found the sum, pair sum and product!

XLatMaths

There are symmetric polynomials in this system of equations
Two of them are elementary symmetric polynomial
and one is sum of powers
For elementary symmetric polynomials we have Vieta formulas
and for sum of powers Newton formulas

holyshit

another method without the use of Vieta's formula:

from first equation, z=1-(x+y)
let m=x+y, n=xy, then z=1-m
now rewrite the second and third equation in terms of m and n

rewrite second equation: xy + = n+m(1-m)=-4

rewrite third equation:
first substitute z=1-m -> x3+y3+(1-m)3=1
note that x3+y3=(x+y)3-3xy(x+y)=m3-3mn
now the equation becomes m3-3mn+(1-m)3=1
note that
hence the entire third equation simplifies to m3-3mn+(1-m)3=1-3m(m-1)-3mn=1

hence 1-3m(m-1)-3mn=1 -> 3m(1-m-n)=0

so, we have either m=0, or 1-m-n=0

if m=0, from the second equtaion it's easy to deduct that n=-4. This gives x+y=m=0 and xy=n=-4, z=1-m=1. It's easy to see that (x, y, z)=(2, -2, 1)

if 1-m-n=0, we have n=1-m. Substitute to second equation eventually gives (m-3)(m+1)=0. This gives additional pairs of possible (m, n) of (3, 2) and (-1, -2). These gives (x, y) pairs of (2, 1) and (-2, 1) respectively - which is simply different permutations of the first answer.

maxm

Oh. Lovely. I used to teach second method to my students. Thanks a lot.

manils

2, -2 and 1, these can be interchanged in x, y, z, and we will obviously have 3! solutions =6

abhinavdiddigam

nice on inspection eq 1 is a plain through {1, 0, 0} let x =1 then y = - z sub into eq 2 -2y = -4 this solves to {1, 2, -2} permutations give all solutions as 3.2 is six...

There is a parity reason to choose 1
its a plain intersecting a sphere which restricts to a circle

you proved it with algbra I can't prove it purely geometricly :(

carlyet

I like the way when you make the solution much easier. Well done

Muslim_

I’ve been trying to solve that on my own before watching this video. After 50 minutes I got a third (imo smoother) method:
As x + y + z equals one (x + y + z)^3 does also equal one.
After a little rearrangement you’ll get
(x + y)³ + 3(x + y)²z + 3(x + y)z² + z³ = 1
After another expantion you’ll get
x³ + y³ + z³ + 3(y + z)x² + 3(y + z)²x + 3(y + z)yz = 1
Now we can subtract the third equation:
3(y + z)x² + 3(y + z)²x + 3(y + z)yz = 0
hence
(y + z)x² + (y + z)²x + (y + z)yz = 0
Now let me devide by (y + z), but in order to do that we have to check whether y + z can equal zero. If it does z = -y
We can substitude that and get:
x = 1
y + -y + -y² = -4
x³ + y³ - y³ = 1

-y² = -4 -> y = ±2; z = -(±2)

Now the other case, y + z ≠ 0:
(y + z)x² + (y + z)²x + (y + z)yz = 0 | divide by (y + z)
x² + (y + z)x + yz = 0
Now we can solve the quadratic with respect to x:
x = -(y + z ±sqrt(y² - 2yz + z²))/2
That’s a perfect square, so
x = -(y + z ±(y - z))/2
And from there you get the two solutions x₁ = -y and x₂ = -z.
But we already solved such an equation. Its solutions were (1 | 2 | -2) and (1 | -2 | 2). Therefore these pairs are the only existing solutions – no matter if you consider complex numbers or not. And because of the symmetry the order of our touple (1 | 2 | -2) does not matter. Thus the set of solutions are all six combinations for 1, 2 and -2,

pseudoexpertise