To multiply like variables with exponents, we will use the following formula:

\(\displaystyle a^x \cdot a^y = a^{x+y}\)

So, we get

\(\displaystyle x^5 \cdot x^2 = x^{5+2}\)

\(\displaystyle x^5 \cdot x^2 = x^7\)

To multiply variables with exponents, we will use the following formula:

\(\displaystyle x^a \cdot x^b = x^{a+b}\)

Now, let’s combine the following:

\(\displaystyle b^4 \cdot b^2 = b^{4+2}\)

\(\displaystyle b^4 \cdot b^2 = b^{6}\)

To multiply like variable with exponents, we will use the following formula:

\(\displaystyle a^x \cdot a^y = a^{x+y}\)

Also, we will multiply coefficients like normal.  

So, we get

\(\displaystyle 3y^3 \cdot 4y^2\)

\(\displaystyle 12(y^3 \cdot y^2)\)

\(\displaystyle 12(y^{3+2})\)

\(\displaystyle 12y^5\)

Simplify the following expression:

\(\displaystyle (9x^5y^2)(12x^3y^8)\)

To combine these, we need to multiply our coefficients and our variables. 

First, multiply the coefficients 

\(\displaystyle 9*12=108\)

Next, multiply our variables by adding the exponent:

\(\displaystyle (x^5y^2)(x^3y^8)=x^{5+3}y^{2+8}=x^8y^{10}\)

So, we put it all together to get:

\(\displaystyle 108x^8y^{10}\)

To divide a variable with an exponent, you can write out all the multiplies, which for this question would look like:

\(\displaystyle \frac{pppppp}{ppp}\)

Then cancel out the matching p's, leaving:

\(\displaystyle ppp=p^{3}\)

Or, if the top and bottom are the same variable, you can simply subtract the bottom exponent from the top exponent.

\(\displaystyle p^{6-3}=p^{3}\)

To divide variables with exponents, either:

Write out the multiplies, then reduce

\(\displaystyle \frac{yy}{yyyyy}=\frac{1}{y^{3}}\)

or subtract the bottom exponent from the top, like so

\(\displaystyle y^{2-5}=y^{-3}=\frac{1}{y^{3}}\)

Remember, negative exponent means dividing. \(\displaystyle y^{-3}\) would also be correct, but you should know that is equivalent to \(\displaystyle \frac{1}{y^{3}}\)

\(\displaystyle \frac{3x(x^2+1)(x+1)(x-1)}{x^3-x}\)

\(\displaystyle =\frac{3x(x^2+1)\left [ (x+1)(x-1) \right ]}{x(x^2-1)}\)

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

\(\displaystyle =\frac{3x(x^2+1)}{x}\)

\(\displaystyle =3(x^2+1)\)

\(\displaystyle \frac{x^6y^2z^3}{x^3y^4z^3}=x^{6-3}y^{2-4}z^{3-3}=x^3y^{-2}z^0=\frac{x^3}{y^2}\)

Start by factoring the numerator. \(\displaystyle x^2y^2\) can be removed from each term.

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

Next, expand the denominator.

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

Simplify by canceling terms.

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

In the denominator, apply the exponent to the terms in parentheses.

\(\displaystyle \frac{2x^{-7}}{(3x)^{-2}}=\frac{2x^{-7}}{(3^{-2}) (x^{-2})}\)

Terms with negative exponents can be inverted within the fraction to eliminate the negative.

\(\displaystyle \frac{2x^{-7}}{(3^{-2}) (x^{-2})}=\frac{2(3^2)(x^2)}{x^7}\)

Simplify by canceling variables and mulitplying.

\(\displaystyle \frac{2(3^2)(x^2)}{x^7}=\frac{18x^2}{x^7}=\frac{18}{x^5}\)