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Given a^2 - 4 = 0, the possible values for a are 2 and -2. Substituting these values into a^3 - 4, we find that a^3 - 4 can be either 4 or -12.

The solution attempts to derive a reduction formula for the integral \(\int_0^\infty \frac{x}{\cosh^n(x)} dx\) using integration by parts and Feynman's trick, but ultimately finds that expressing the integral in terms of hypergeometric functions is a more suitable approach.

The integral \int_{0}^{\infty} \frac{x}{(\cosh x)^2} dx evaluates to ln(2).

To evaluate the limit, direct substitution results in an indeterminate form (0/0). Factoring the numerator reveals a common factor of (x-3), which cancels with the denominator. Substituting x=3 into the simplified expression yields a final answer of 58.

The eigenvalues of the matrix [[3, 1], [1, 3]] are 4 and 2.

The eigenvalues of the matrix [[0, 1], [-1, 0]] are i and -i.

The eigenvalues of the given 3x3 matrix are 1, 2, and 2.

The problem involves finding the equation of a circle given a point on the circle, the radius, and the fact that the center of the circle lies on a given line. Two possible circles satisfy the conditions, each with a different center, and the equations for both circles are derived.

The solution finds the trace lines of a plane by setting one coordinate to zero and solving for the other two coordinates in terms of the remaining parameter. This results in parametric equations for the trace lines in the xy, xz, and yz planes.

The equation represents an ellipse centered at (-2, 0) with a vertical major axis of length 4 and a horizontal minor axis of length 2.