Question

$\frac{(3a^2b)^3}{9(ab)^4}$

Answer 1

You first multiply the outside exponents into the numbers in the parentheses.

When you have an exponent being multiplied directly to another exponent, you multiply the exponents together.

For example(because I am a bad explainer):

$(x^{2} )^4= x^{2(4)} = x^8$

$(x^4)^3 = x^{4(3)} = x^{12}$

When you divide an exponent by an exponent, you subtract the exponents

For example:

$\frac{x^4}{x^1} =x^{4-1}=x^3$

When you have a negative exponent, you move it to the other side of the fraction to make the exponent positive

For example:

$x^{-3}=\frac{1}{x^3}$

$\frac{1}{y^{-5}}=\frac{y^5}{1}=y^5$

$\frac{(3a^2b)^3}{9(ab)^4}$

You can think of it like this if you want:

$\frac{(3^1a^2b^1)^3}{9(a^1b^1)^4}$  Now multiply the outside exponents into the exponents in the parentheses

$\frac{3^3a^6b^3}{9(a^4b^4)} =\frac{27a^6b^3}{9(a^4b^4)}$ Divide 27 and 9

$\frac{3a^6b^3}{a^4b^4} =(3)(a^{6-4})(b^{3-4})=(3)(a^2)(b^{-1})=(3)(a^2)(\frac{1}{b^1})=\frac{3a^2}{b}$

Your answer is $\frac{3a^2}{b}$