Question

(x2 + y2 − 1)3 − x2y3 = 0​

Answer 1

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Answer 2

The equation you provided is a polynomial equation in two variables, x and y. It appears to be a cubic equation. Let's simplify it and analyze it further.

The equation is:

(x^2 + y^2 - 1)^3 - x^2y^3 = 0

Expanding the cube on the left side, we get:

(x^2 + y^2 - 1)(x^2 + y^2 - 1)(x^2 + y^2 - 1) - x^2y^3 = 0

Now, we can expand this further:

(x^2 + y^2 - 1)(x^4 + 2x^2y^2 + y^4 - 2x^2 - 2y^2 + 1) - x^2y^3 = 0

Expanding again, we have:

x^6 + 2x^4y^2 + x^2y^4 + y^6 - 2x^4 - 4x^2y^2 - 2y^4 + x^2 - 2y^2 + 1 - x^2y^3 = 0

Rearranging terms, we get:

x^6 + y^6 + x^2y^4 - x^2y^3 + 2x^4y^2 - 4x^2y^2 - 2y^4 + x^2 - 2y^2 + 1 = 0

Simplifying further, we have:

x^6 + y^6 + x^2(y^4 - y^3) + 2x^2y^2(x^2 - 2) - 2(y^4 + y^2) + (x^2 - 2y^2 + 1) = 0

At this point, it is clear that this equation is a polynomial of degree 6 in x and y, with various terms involving powers of x and y. It does not have a simple, closed-form solution. If you have specific values for x and y or if you need to solve this equation numerically, you can use numerical methods such as Newton's method or other root-finding algorithms to approximate the solutions.