To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so add the coefficients and retain the base and exponent:

-8c² + 6c²
(-8 + 6)c²
-2c²

The distributive property for division helps in solving expressions like \({b + c \over a}\). It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: \({b + c \over a} = {b \over a} + {c \over a}\). For example, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).

To add these fractions, first find the lowest common multiple of their denominators. The first few multiples of 4 are [4, 8, 12, 16, 20, 24, 28, 32, 36, 40] and the first few multiples of 8 are [8, 16, 24, 32, 40, 48, 56, 64, 72, 80]. The first few multiples they share are [8, 16, 24, 32, 40] making 8 the smallest multiple 4 and 8 share.

Next, convert the fractions so each denominator equals the lowest common multiple:

\( \frac{5 x 2}{4 x 2} \) + \( \frac{3 x 1}{8 x 1} \)

\( \frac{10}{8} \) + \( \frac{3}{8} \)

Now, because the fractions share a common denominator, you can add them:

\( \frac{10 + 3}{8} \) = \( \frac{13}{8} \) = 1\(\frac{5}{8}\)

The greatest common factor (GCF) is the greatest factor that divides two integers.

A factorial is the product of an integer and all the positive integers below it. To solve a fraction featuring factorials, expand the factorials and cancel out like numbers:

\( \frac{3!}{6!} \)
\( \frac{3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \)
\( \frac{1}{6 \times 5 \times 4} \)
\( \frac{1}{120} \)