Question

(2^2b^-2)^-3 Math help plz

Answer 1

Answer:

$\dfrac{b^6}{64}$

Step-by-step explanation:

We need three rules. Raising a power to a power. Raising a product to a power. Raising a number to a negative exponent.

Rule 1:

To raise a power to a power, multiply powers.

$(a^m)^n = a^{mn}$

Rule 2:

To raise a product to a power, raise every factor of the product to the power.

$(a^xb^yc^z)^n = a^{xn}b^{yn}c^{zn}$

Rule 3:

To raise a number to a negative power, follow this formula.

$a^{-n} = \dfrac{1}{a^n}$

Your problem.

$(2^2b^{-2})^{-3} =$

You have a product raised to a power, so raise each factor to the power.

$= (2^2)^{-3}(b^{-2})^{-3}$

Now raise each power to a power by multiplying exponents.

$= 2^{2 \times (-3)}b^{-2 \times (-3)}$

$= 2^{-6}b^{6}$

Now we follow the rule of a negative exponent.

$= \dfrac{1}{2^6} \times b^6$

$= \dfrac{b^6}{2^6}$

$= \dfrac{b^6}{2 \times 2 \times 2 \times 2 \times 2 \times 2}$

$= \dfrac{b^6}{64}$

Answer: $\dfrac{b^6}{64}$