Dividing by 3 gives x² - 4x + 4 = 0, which factors to (x - 2)² = 0, hence the root is 2.

This is a perfect square: (2x - 3)² = 0, hence the root is x = 1.5.

Factoring gives (x + 4)(x - 2) = 0, hence the roots are 4 and -2.

f'(x) = 6x^2 - 6. f'(1) = 6(1)^2 - 6 = 0.

f'(x) = 2x + 2. At x = 1, f'(1) = 4.

This is a separable equation. Integrating gives y = 1/(C - x).

This is a linear first-order equation. The integrating factor is e^(3x). The solution is y = Ce^(3x) + 2.

This is a separable equation. The solution is y = tan(x + C).

This is a first-order linear differential equation. The solution is y = Ce^(-2x).

The sum of the first n terms of a geometric series is S_n = a(1 - r^n) / (1 - r). Here, a = 2, r = 3, n = 15. So, S_15 = 2(1 - 3^15) / (1 - 3) = 2(1 - 14348907) / -2 = 14348906.

The series is the sum of squares: 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 1 + 4 + 9 + 16 + 25 = 55.

The term containing x^3 is C(6,3) * (5)^3 = 20 * 125 = 250.

The term containing x^3 is C(5,3) * x^3 * (-1)^2 = 10 * x^3 * 1 = 10.

The term independent of x occurs when the powers of x cancel out. The term is 81.

The term independent of x occurs when the powers of x cancel out. The term is C(5,2)(-3)^2(1)^3 = 45.

The expression can be rewritten as (x - 2)^4. The term independent of x occurs when k = 4, which gives us (-2)^4 = 16.

The expression can be rewritten as (x - 2)^6. The term independent of x occurs when k = 3, which gives us 6C3 * (-2)^3 = 20 * (-8) = -160. The term independent of x is 24.

Using the binomial theorem, (3 + 2)^3 = C(3,0) * 3^3 * 2^0 + C(3,1) * 3^2 * 2^1 + C(3,2) * 3^1 * 2^2 + C(3,3) * 3^0 * 2^3 = 27 + 54 + 36 + 8 = 125.

3^3 = 27 and 2^3 = 8, so 27 - 8 = 19.

5! = 5 × 4 × 3 × 2 × 1 = 120.

5^3 = 5 × 5 × 5 = 125.

9 × 9 = 81 and 3 × 3 = 9, so 81 - 9 = 72.

9 × 9 = 81 and 5 × 5 = 25, so 81 - 25 = 56.

9 × 9 = 81 and 5 × 5 = 25, so 81 - 25 = 56.

9 × 9 = 81, then 81 - 7 = 74.

The discriminant must be negative: 4² - 4*1*k < 0, which gives k > 4, so the minimum value is -6.

For equal roots, the discriminant must be zero: k² - 4*1*16 = 0, thus k² = 64, k = ±8. The value of k can be -8.

For no real roots, the discriminant must be negative: k² - 4*1*9 < 0, thus k² < 36, hence k < -6 or k > 6.

For no real roots, the discriminant must be less than zero: k² - 4*1*16 < 0, which gives k > 8.

Setting the determinant to zero: \( 1*x - 2*3 = 0 \) gives \( x - 6 = 0 \) or \( x = 6 \).