Find an approximation of $(0.99)^{5}$ using the first three terms of its binomial expansion.

The coefficient of $x^2$ in the expansion of the product $(2 - x^2)((1 + 2x + 3x^2)^6 + (1 - 4x^2)^6)$ is

The sum of all the coefficients in the binomial expansion of $(x^2 + x - 3)^{319}$ is

$(\sqrt{2} + 1)^6 - (\sqrt{2} - 1)^6 = $

Find the value of $\frac{18^3 + 7^3 + 3 \times 18 \times 7 \times 25}{3^6 + 6 \times 243 \times 2 + 15 \times 81 \times 4 + 20 \times 27 \times 8 + 15 \times 9 \times 16 + 6 \times 3 \times 32 + 64}$

If $n$ is a positive integer,then $(\sqrt{3} + 1)^{2n} - (\sqrt{3} - 1)^{2n}$ is