Solution:

Using the identity:
$$a^2 – b^2 = (a – b)(a + b)$$
Substituting values:
$$15 = 3 \times (a + b)$$
$$a + b = \frac{15}{3} = 6$$
$$\boxed{6}$$

Solution:

Using the identity:
$$A^2 – B^2 = (A – B)(A + B)$$
Setting $A = x + 2y$ and $B = x – 2y$:
$$(x + 2y)^2 – (x – 2y)^2 = [(x + 2y) – (x – 2y)][(x + 2y) + (x – 2y)]$$
$$= (x + 2y – x + 2y)(x + 2y + x – 2y)$$
$$= (4y)(2x) = 8xy$$
$$\boxed{8xy}$$

Solution:

Using the identity:
$$(a + b)(a – b) = a^2 – b^2$$
Adding $b^2$:
$$a^2 – b^2 + b^2 = a^2$$
$$\boxed{a^2 + b^2}$$

Solution:

Using the identity:
$$x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 – 2$$
Substituting $x + \frac{1}{x} = 4$:
$$x^2 + \frac{1}{x^2} = 4^2 – 2 = 16 – 2 = 14$$
$$\boxed{14}$$

Solution:

Using the identity:
$$p^2 + q^2 = (p + q)^2 – 2pq$$
Substituting values:
$$p^2 + q^2 = 7^2 – 2(10) = 49 – 20 = 39$$
$$\boxed{39}$$

Solution:

Recognizing the identity:
$$a^3 + 3a^2b + 3ab^2 + b^3 = (a + b)^3$$
Substituting $a = x$ and $b = y$:
$$x^3 + 3x^2y + 3xy^2 + y^3 = (x + y)^3$$
$$\boxed{(x + y)^3}$$

Solution:

Using the identity:
$$p^3 + q^3 + r^3 – 3pqr = (p + q + r)(p^2 + q^2 + r^2 – pq – qr – rp)$$
For $p = a – b$, $q = b – c$, and $r = c – a$:
$$(a – b) + (b – c) + (c – a) = 0$$
Since the sum is zero, the entire expression evaluates to:
$$\boxed{0}$$

Solution:

Using the identity:
$$x^3 + y^3 = (x + y)(x^2 – xy + y^2)$$
We first find:
$$x^2 + y^2 = (x + y)^2 – 2xy = 10^2 – 2(21) = 100 – 42 = 58$$
Now:
$$x^3 + y^3 = 10(58 – 21) = 10(37) = 370$$
$$\boxed{1000}$$

Solution:

Given equations:
$$2x + 3y = 12$$
$$4x – 6y = 18$$
Dividing the second equation by 2:
$$2x – 3y = 9$$
Adding both equations:
$$(2x + 3y) + (2x – 3y) = 12 + 9$$
$$4x = 12 + 9 = 21$$
$$x = \frac{21}{4} = 3$$
$$\boxed{3}$$

Solution:

Using the quadratic formula:
$$x = \frac{-(-8) \pm \sqrt{(-8)^2 – 4(3)(4)}}{2(3)}$$
$$x = \frac{8 \pm \sqrt{64 – 48}}{6}$$
$$x = \frac{8 \pm \sqrt{16}}{6}$$
$$x = \frac{8 \pm 4}{6}$$
$$x = \frac{8 + 4}{6} = \frac{12}{6} = 2, \quad x = \frac{8 – 4}{6} = \frac{4}{6} = \frac{2}{3}$$
$$\boxed{\frac{4}{3}}$$

Solution:

Using the Remainder Theorem, the remainder when $f(x)$ is divided by $x – 1$ is found by evaluating $f(1)$:

$$f(x) = x^3 + 3x + 2$$

Substituting $x = 1$:

$$1^3 + 3(1) + 2 = 1 + 3 + 2 = 6$$

$$\boxed{6}$$

Solution:

Using the identity:
$$a^3 + b^3 = (a + b)(a^2 – ab + b^2)$$

First, calculate $a^2 + b^2$:

$$a^2 + b^2 = (a + b)^2 – 2ab = 5^2 – 2(6) = 25 – 12 = 13$$

Now:

$$a^3 + b^3 = (5)(13 – 6) = 5(7) = 75$$

$$\boxed{75}$$

Solution:

Rewriting:

$$(x – 3)(x – 3) = 0$$

$$(x – 3)^2 = 0$$

$$x = 3$$

$$\boxed{3}$$

Solution:

Factoring:

$$x^2 – 5x + 6 = (x – 2)(x – 3) = 0$$

Setting each factor to zero:

$$x – 2 = 0 \quad \Rightarrow \quad x = 2$$

$$x – 3 = 0 \quad \Rightarrow \quad x = 3$$

$$\boxed{2, 3}$$

Solution:

We need to factor:

$$x^2 – 4x – 12$$

Find two numbers whose product is $-12$ and sum is $-4$:

$$(-6, +2)$$

$$x^2 – 4x – 12 = (x – 6)(x + 2)$$

$$\boxed{(x – 6)(x + 2)}$$