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 \(-4\):

\[
(-6, +2)
\]

\[
x^2 – 4x – 12 = (x – 6)(x + 2)
\]

\[
\boxed{(x – 6)(x + 2)}
\]