Never divide fractions!  Simply flip the fraction that follows the division symbol, and then multiply it by the first fraction.  So this expression becomes:

\(\displaystyle \dpi{100} \frac{1}{3}\times \frac{9}{1}=\frac{9}{3}=3\)

\(\displaystyle \frac{6x^3-3x^2+9x}{3x}=\frac{6x^3}{3x}-\frac{-3x^2}{3x}+\frac{9x}{3}=2x^2-x+3\)

\(\displaystyle \frac{4x^{4}- 8 x^{2} - 16} {2x}\)

\(\displaystyle = \frac{4x^{4}} {2x} - \frac{ 8 x^{2}} {2x} - \frac{16} {2x}\)

\(\displaystyle = \frac{2x^{4-1}} {1} - \frac{ 4 x^{2-1}} {1} - \frac{8} {x}\)

\(\displaystyle = \frac{2x^{3}} {1} - \frac{ 4 x^{1}} {1} - \frac{8} {x}\)

\(\displaystyle = 2x^{3} - 4 x - \frac{8} {x}\)

\(\displaystyle \frac{8x^4+4x^3-16x}{4x}\)

\(\displaystyle =\frac{8x^4}{4x}+\frac{4x^3}{4x}-\frac{16x}{4x}\)

\(\displaystyle =2x^3+x^2-4\)

In this division problem, you can simplify first the coefficients, then the variables.

Each of the coefficients in the numerator is divisible by the coefficient in the denominator, allowing you to divide out and cancel the 4 in the denominator:

\(\displaystyle \frac{16y^3 - 4y^2 + 8y}{4y} = \frac{4y^3 - y^2 + 2y}{y}\)

Finally, you can simplify the varaibles. Remember that when simplifying variables in a fraction (division problem), you subtract the numerator variable's exponent by the denominator variable's exponent. You can do this with each term in this problem, because each term in the numerator has at least a \(\displaystyle y\):

\(\displaystyle \frac{4y^3 - y^2 + 2y}{y} = 4y^2 - y +2\)

In this division problem, you can simplify first the coefficients, then the variables.

Each of the coefficients in the numerator is divisible by the coefficient in the denominator, allowing you to divide out and cancel the 7 in the denominator:

\(\displaystyle \frac{7x^4 + 21x^2 - 14}{7x^2} = \frac{x^4 + 3x^2 - 2}{x^2}\)

Finally, you can simplify the varaibles. Remember that when simplifying variables in a fraction (division problem), you subtract the numerator variable's exponent by the denominator variable's exponent.

You can only do this with the first two terms in the numerator, since the final term does not have a variable. Instead, the final term will keep \(\displaystyle x^2\) in the denominator:

\(\displaystyle \frac{x^4 + 3x^2 - 2}{x^2} = x^2 + 3 - \frac{2}{x^2}\)

\(\displaystyle \frac{x^2+2xy+xy^2}{x}=\frac{x^2}{x}+\frac{2xy}{x}+\frac{xy^2}{x}=x+2y+y^2\)

The first step to solving this question is to reduce \(\displaystyle (8-2)^{2}\). 

\(\displaystyle (8-2)^{2}\)

\(\displaystyle 6^{2}\)

\(\displaystyle 36\)

The only number listed that is a factor of 36 is 18, given that 2 times 18 is 36. Therefore, 18 is the correct answer. 

To divide this problem we simplify it first. In this problem we can separate the big fraction into three smaller fractions. 

\(\displaystyle \frac{4x^{3}+8x^{2}-12x}{4x}=\frac{4x^{3}}{4x}+\frac{8x^{2}}{4x}-\frac{12x}{4x}\)

Then from here, we can pull out a \(\displaystyle 4x\) from both the numerator and denominator of each smaller fraction.

\(\displaystyle \frac{4x(x^2)}{4x}+\frac{4x(2x)}{4x} - \frac{4x(3)}{4x}\)

Now we cancel terms and get the following result:

\(\displaystyle x^2+2x-3\)

To simplify this problem we first separate the large fraction into three smaller fractions.

\(\displaystyle \frac{2x^{4}+6x^{3}+8}{2x}\)

\(\displaystyle =\frac{2x^{4}}{2x}+\frac{6x^{3}}{2x}+\frac{8}{2x}\)

From here we can factor out \(\displaystyle 2x\) from the numerator and denominator of the first two fractions.

\(\displaystyle = \frac{2x(x^3)}{2x}+\frac{2x(3x^2)}{2x}+\frac{8}{2x}\)

The \(\displaystyle 2x\) can be canceled out in the first two fractions. From here we can factor out a \(\displaystyle 2\) from the numerator and denominator of the third fraction.

\(\displaystyle =x^3+3x^2+\frac{2(4)}{2(x)}\)

Thus becoming: 

\(\displaystyle =x^{3}+3x^{2}+\frac{4}{x}\)