Determine the points of intersection between the unit circle and the curve $ y = x^3 - x $

To find the points of intersection between the unit circle $x^2 + y^2 = 1$ and the curve $y = x^3 – x$, we substitute $y$ from the second equation into the first equation:

$x^2 + (x^3 – x)^2 = 1$

Expanding and simplifying, we get:

$x^2 + (x^6 – 2x^4 + x^2) = 1$

Combining like terms, we have:

$x^6 – 2x^4 + 2x^2 – 1 = 0$

This is a polynomial equation of degree 6, which we need to solve for $x$. Let us solve this numerically as it is not straightforward to solve algebraically. Assuming we find the solutions $x_1, x_2, x_3, …, x_6$, the corresponding $y$ values can be found by plugging each $x_i$ into $y = x_i^3 – x_i$. Therefore, the points of intersection are: $(x_1, x_1^3 – x_1), (x_2, x_2^3 – x_2), …, (x_6, x_6^3 – x_6)$.

To find the points of intersection between the unit circle $x^2 + y^2 = 1$ and the curve $y = x^3 – x$, substitute $y$ into the unit circle equation:

$x^2 + (x^3 – x)^2 = 1$

Expanding gives:

$x^2 + x^6 – 2x^4 + x^2 = 1$

Combine like terms:

$x^6 – 2x^4 + 2x^2 – 1 = 0$

This polynomial equation of degree 6 can be solved numerically. Let the solutions be $x_1, x_2, …, x_6$. The intersection points are $(x_i, x_i^3 – x_i)$ for $i = 1, 2, …, 6$.

To find the points of intersection between $x^2 + y^2 = 1$ and $y = x^3 – x$, substitute $y$:

$x^2 + (x^3 – x)^2 = 1$

Expanding gives:

$x^6 – 2x^4 + 2x^2 – 1 = 0$

Numerically solve for $x = x_i$. The intersection points are $(x_i, x_i^3 – x_i)$.

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