3-x2y3%3D0-solution.jpeg "(x2+y2-1)3-x2y3=0 Solution")
Solving the Equation (x^2 + y^2 - 1)^3 - x^2y^3 = 0
In this article, we will explore the solution to the equation (x^2 + y^2 - 1)^3 - x^2y^3 = 0. This equation may seem complex, but with the right approach, we can find the solutions.
Step 1: Simplify the Equation
Let's start by simplifying the equation. We can begin by expanding the cube:
(x^2 + y^2 - 1)^3 = x^6 + 3x^4y^2 + 3x^2y^4 - x^4 - 3x^2y^2 - y^4 + 1
Now, subtract x^2y^3 from both sides of the equation:
x^6 + 3x^4y^2 + 3x^2y^4 - x^4 - 3x^2y^2 - y^4 + 1 - x^2y^3 = 0
Step 2: Factorize the Equation
Next, we'll try to factorize the equation. After some manipulation, we get:
(x^2 + y^2 - 1)(x^4 - x^2y^2 - y^4 + 1) - x^2y^3 = 0
Step 3: Analyze the Factors
Now, we'll analyze the factors. The first factor x^2 + y^2 - 1 equals to zero when x^2 + y^2 = 1, which is the equation of a unit circle centered at the origin.
The second factor x^4 - x^2y^2 - y^4 + 1 can be rewritten as (x^2 - 1)(x^2 - y^2). This factor equals to zero when x^2 = 1 or x^2 = y^2.
Step 4: Find the Solutions
Combining the results from Steps 2 and 3, we get the following solutions:
x^2 + y^2 = 1(unit circle)x^2 = 1(two vertical lines:x = 1andx = -1)x^2 = y^2(two diagonal lines:y = xandy = -x)
These are the solutions to the equation (x^2 + y^2 - 1)^3 - x^2y^3 = 0.
Conclusion
In this article, we successfully solved the equation (x^2 + y^2 - 1)^3 - x^2y^3 = 0. The solutions include a unit circle, two vertical lines, and two diagonal lines. This problem demonstrates the importance of algebraic manipulation and factorization in solving complex equations.
