Odd result when considering x³ - y³ = 91

Question

By inspection one may notice that x = 3, y = -4 is a solution & so must be x = 4, y = -3. If this equation defines a cubic, then with 2 real roots there must be 1 more real solution. To find it factor the difference of two cubes as (x - y)(x² + xy + y²) = 91. Now it is easy to note that (x - y) = 7 for the two established roots. So we are dealing with (x² + xy + y²) = 13. The symmetry suggests that x = y = w & 3w² = 13 so we also have x = y = ±√(13/3). But now we have a cubic with 4 roots. Which is a dilemma. Wolfram alpha considers this (equation in Title) to be an elliptic curve with only one root, said root not agreeing with any of my four. (Wolfram alpha has a bug.) Wolfram alpha similarly considers (x² + xy + y²) = 13 to be an ellipse with roots that do not have the property x = y. They also don't agree with the graph Wolfram alpha produced for the ellipse. (More bugs). If we substitute u = y/2, then the quadratic factor equation can be written (x + u)² + (3u² - 13) = 0. Wolfram alpha seems to have used this form when solving for (x,y). Thank you very much for your help. Yours, Dane

Answer 1

x - y = 1 and x² + xy + y² = 91; y = x - 1; x² + x² - x + x² - 2x + 1 = 91; x² - x - 30 = 0; x = 6 or x = - 5

and we get 2 solutions (6, 5), (- 5, - 6).

Let name x - y = u, x² + xy + y² = v.

Then possible such cases else: u = 91; v = 1; u = - 1; v = - 91; u = - 91; v = - 1

u = 7; v = 13, u = 13, v = 7

u = - 7; v = - 13; u = - 13; v = - m7.

In each we solve quadratic equation and take integer solution.

Answer 2

Hang on… I think we’re mixing some things here.

Yes, we have a “cubic”, a cubic in 2 variables… f(x, y) = x³ - y³ - 91 = 0

However, the idea that we are limited to 3 solutions, and we find at least 4, is not what The Fundamental Theorem of Algebra (d’Alembert’s Theorem) speaks to. The FTA applies to single-variable polynomials Pₙ(x) = 0 with complex (or real) coefficients, which will have exactly ’n’ roots, complex and/or real.

Yes, y = x³ - 91 = x³ - (∛91)³ will have 3 roots… 1 root at ∛91 ==>

y = (x - ∛91)(x² + (∛91)x + (91)²/³) = (x - ∛91)(x + (∛91 / 2)[1 - i√3])(x + (∛91 / 2)[1 + i√3])

As for f(x, y) = x³ - y³ - 91 = 0, it is true that (3, -4) and (4, -3) are solutions to f(x, y), but the solutions are infinite.

To see this, consider dy/dx (via implicit differentiation)…

3x² - 3y²(dy/dx) = 0 ==> dy/dx = x²/y² ≥ 0 ∀ (x, y) in f()

This means that the entire curve for y = f(x) is monotonically increasing, with something “funky” at y=0, and probably at x=0. The monotonicity also implies that for _every_ conceivable ‘x’ ∈ ℝ, there is a unique corresponding ‘y’, also ∈ ℝ… the function f(x, y) is 1-to-1 from (-∞, ∞).

We can go further to look for inflection points: d²/dx² y = d/dx (x²/y²) = (2xy³ - 2x⁴) / y⁵

There is still funkiness at y=0, but that is addressed by studying when...

2xy³ - 2x⁴ = 0 ==> x = 0 _OR_ y³ = x³. (x, y=x) are not points anywhere on f(x,y), so we are left with something interesting at x=0, and perhaps another inflection point at y=0 where d²y/dx² is undefined, so we investigate.

At x=0, we have y = -∛91 ≈ -4.498..., and at y=0, we have x=∛91 ≈ 4.498, and looking at the following graph confirms those 2 inflection points… (0, -∛91) and (∛91, 0)

Anyway, I plugged in f(x, y) in Desmos, and see nothing unusual.

I suspect this just a misapplication of the Fundamental Theorem of Algebra, and expectations thereof ? Well, good reminder in any case !

Answer 3

x^3 - y^3 = 91

125 -34 = 91

x=5, y = cube root of 34

64 +37 = 91

x =4, y =cube root of -37

an infinite number of similar irrational solutions

27 + 64 = 91

x=3, y = -4 an integer solution, which is what you're likely looking for

216 -125 = 91

x =6, y =5 another integer solution

you can graph it and look for x intercepts but about all that does is give you y=0, x=cube root of 91 = slightly less than 4.5. Still look along the curve and find where it seems to intersect vertex of a square, such as the point (6,5) x=6 y=5 is another solution

Answer 4

If I was determining the roots I would rewrite the function as:

f(x) = (x³ - 91)^(1/3)

f(x) = (x - 91^1/3)^1/3 (x² + 91^1/3x + 91^2/3)^1/3

The roots exist when:

x = 91^1/3 or let k = 91^1/3, x = [-k ± √(k² - 4(1)(k²))]/2 = [-k ± ki√3]/2

This Desmos graph has complex mode turned on and shows that evaluating the function at those 3 roots does give a value of zero.

desmos.com/calculator/pva1rixtgy

However, evaluating f(4) gives a strange result with complex mode turned on. Turn complex mode off and the result for f(4) = -3. I have submitted a support request to Desmos asking if this is a bug in complex mode.