An Interesting Polynomial Equation

[@Blaqjaqshellaq]

Consider the ratio of x to y: If x=y, x/y=1 If x=0 y=i, so x/y=0 If x=i*(4/3)^1/2 y=i*(1/3)^1/2, so x/y=1/2.

[@johns.8246]

y^3 - y = x^3 - x is way more interesting. Try that one instead.

[@rickwilliams9001]

Couldn’t you sat x=y by inspection.

[@dan-florinchereches4892]

Yes but there may be other solutions. If you use a^3-b^3 identity you get (y-x)(y^2+xy+x^2+1)=0 The second equality is almost a perfect square so by multiplying by 2 and grouping X^2+2xy+y^3+x^2+y^2+2=0 (X+y)^2+x^2+y^2+2>=2 because a square is >=0 Now we know for sure y=x is the only solution. Otherwise there would be some relations between X and y for which more solutions exist

[@brendanward2991]

I immediately saw that y=x is one solution, so y-x is a factor. Divide the polynomial by y-x to find the quadratic factor, and solve using the quadratic formula.

[@SyberMath]

Good point!

[@aliaims760]

@@SyberMathI said x = y is an obvious solution and derivated the function x |-> x^3 +x to show that it’s Strictly monotonous and continuous therefore there can only be one solution which is the one we found. Does that work?

[@fahrenheit2101]

Had to solve this the other day as part of a longer q Took me a while, but once you realize, it's very trivial. Monotonic function, invertible, x = y Again... Edit: I suppose that's only in the reals tbf

[@Don-Ensley]

problem y³ +y = x³ +x Solving for y values. Standard form for cubic equation is y³ + y - (x³ +x) = 0 By cubic formula, y = a+b for solution y³ + y = (x³ +x) -3ab = 1 a³ +b³ = (x³ +x) a³ b³ = -1/27 b³ = -1/(27a³) a³ -1/(27a³) = (x³ +x) ( a³ )² -(x³ +x)( a³ ) - 1/27 = 0 a³ = { x³ +x ± √[(x³ +x)²+4/27] } / 2 a³ +b³ = (x³ +x) b³ = x³ +x - a³ = x³ +x - { x³ +x ± √[(x³ +x)²+4/27] } / 2 = { 2 x³ + 2 x - x³ - x ∓ √[(x³ +x)²+4/27] } / 2 = { x³ + x ∓ √[(x³ +x)²+4/27] } / 2 a = ∛ ( { x³ +x ± √[(x³ +x)²+4/27] } / 2 ) b = ∛( { x³ + x ∓ √[(x³ +x)²+4/27] } / 2 ) y = a + b y = ∛ ( { x³ +x + √[(x³ +x)²+4/27] } / 2 ) + ∛( { x³ + x - √[(x³ +x)²+4/27] } / 2 ) answer y = ∛ ( { x³ +x + √[(x³ +x)²+4/27] } / 2 ) + ∛( { x³ + x - √[(x³ +x)²+4/27] } / 2 )

[@vladimirkaplun5774]

???????? y^3-x^3=x-y ->(y-x)(x^2+xy+y^2+1)=0->x=y Full stop in 20 sec

[@WahranRai]

Same kind as already proposed : kzbin.info/www/bejne/mn-apYCJl8RmkM0

[@scottleung9587]

I used method 2b.

[@SyberMath]

2b or not 2b? 😜😁

[@scottleung9587]

@@SyberMath exactly!

[@roberttelarket4934]

x = y.

[@rakenzarnsworld2]

x = y bruh

[@SyberMath]

nice