(x - y)³ = x³ - 3x²y + 3xy² - y³ by the binomial theorem.

a² - b² = (a + b)(a - b) by the difference of squares identity.

(a - b)(a + b) = a² - b² by the difference of squares formula.

(a - b)² = a² - 2ab + b² by the expansion of the square of a binomial.

(x + y + z)³ = x³ + y³ + z³ + 3xyz.

(x + y)³ = x³ + y³ + 3xy(x + y) by the binomial expansion.

(x + 2)² = x² + 2*2*x + 2² = x² + 4x + 4.

The smallest positive integer is 1.

log_10(0.1) = log_10(10^-1) = -1.

log_a(b) + log_a(c) = log_a(bc) by the product property of logarithms.

If a > 1, then log_a(b) < log_a(c) when b < c.

log_a(bc) = log_a(b) + log_a(c) by the product rule of logarithms.

(a + b)³ = a³ + 3a²b + 3ab² + b³.

(x + y)² = x² + 2xy + y².

Parallel lines have the same slope. The slope of the given line is 3/4, and the line 6x - 8y + 10 = 0 also has the same slope.

A diagonal matrix is one where all off-diagonal elements are zero. The matrix [[1, 0], [0, 2]] is a diagonal matrix.

A zero matrix is one where all elements are zero. The matrix [[0, 0], [0, 0]] meets this criterion.

An identity matrix has 1s on the diagonal and 0s elsewhere. The matrix [[1, 0], [0, 1]] fits this definition.

An orthogonal matrix is defined as a matrix whose transpose is equal to its inverse.

A matrix is not invertible if its determinant is zero. The matrix [[0, 1], [0, 0]] has a determinant of 0.

A symmetric matrix is one that is equal to its transpose. Matrix B is symmetric because B = B^T.

25 is a perfect square (5^2).

Substituting x = 1 into the equation gives y = 1^2 - 4 = -3, so the point (1, -3) lies on the parabola.

To check if a point lies on the parabola, substitute the x-coordinate into the equation. For (2, 4), 4^2 = 16 and 8*2 = 16, so it lies on the parabola.

The polar form of a complex number r(cos θ + i sin θ) where r is the modulus and θ is the argument. Here, 4 = 4(cos 0 + i sin 0).

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.

The function f(x) = 1/(x-1) is not continuous at x = 1 because it is undefined there.

The function f(x) = 1/(x-3) is not defined at x = 3, hence it is not continuous at that point.

The function f(x) = |x| is continuous everywhere, including at x = 0.

Variance is always non-negative as it is calculated as the average of squared deviations from the mean.