Solve (-4/9)^-3: Negative Exponent with Fractions

First, we’ll apply the exponent law for a power raised over a product of terms:

$(x\cdot y)^n=x^n\cdot y^n$

Let's apply this law to the problem:

$\big(-\frac{4}{9}\big)^{-3}= \big(-1\cdot \frac{4}{9}\big)^{-3}=(-1)^{-3}\cdot\big( \frac{4}{9}\big)^{-3}$

Next, we'll recall the exponent law for negative exponents:

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

Next, we’ll simplify the left-hand factor from the product we obtained in the previous step:

$(-1)^{-3}=\frac{1}{(-1)^3}=\frac{1}{-1}=-1$

From here we'll handle the right term in the multiplication from the expression we got in the last step:

$\big( \frac{4}{9}\big)^{-3}$

Let's recall again the exponent law for negative exponents that we mentioned earlier:

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

However, before we continue handling the above term, let's understand this law in a slightly different, indirect way:

Let's note that if we treat this law as an equation (and it is indeed an equation in every sense), and multiply both sides of the equation by the common denominator which is:

$a^n$ we get:

$a^{-n}=\frac{1}{a^n}\\ \frac{a^n}{1} =\frac{1}{a^n}\hspace{8pt} \text{/}\cdot a^n\\ a^n\cdot a^{-n}=1$

When in the first stage we remembered that any number can be represented as itself divided by 1, we applied this to the left side of the equation, then multiplied by the common denominator and to know by how much we multiplied each numerator (after reduction with the common denominator) we addressed the question "By how much did we multiply the current denominator to get the common denominator?".

Let's look at the result we got:

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

This means that $a^n,\hspace{4pt}a^{-n}$are reciprocal numbers, or in other words:

$a^n$ is reciprocal to $a^{-n}$ (and vice versa),

And specifically:

$a,\hspace{4pt}a^{-1}$ are reciprocal to each other,

We can apply this understanding to the problem if we also remember that the reciprocal of a fraction is obtained by switching the numerator and denominator, meaning that the fractions:

$\frac{a}{b},\hspace{4pt}\frac{b}{a}$

are reciprocal fractions - which is easily understood, since their multiplication clearly gives the result 1,

And if we combine this with our previous understanding, we can easily conclude that:

$\big(\frac{a}{b}\big)^{-1}=\frac{b}{a}$

This means that raising a fraction to the power of negative one will give a result that is the reciprocal fraction, obtained by switching the numerator and denominator.

Let's return to the problem and apply these understandings, additionally recalling the exponent law for exponents of exponents, but in the opposite direction:

$(a^m)^n=a^{m\cdot n}$

Let's apply this law to the problem for the right term in the multiplication in the last expression we got, and we get:

$\big( \frac{4}{9}\big)^{-3} = \big(\frac{4}{9}\big)^{-1\cdot3}= \big(\big(\frac{4}{9}\big)^{-1}\big)^{3} =\big(\frac{9}{4}\big)^{3}$

When in the first stage we expressed the exponent as a multiplication between two numbers, in the second stage we applied the above exponent law for exponents of exponents in the opposite direction and in the next stage we applied within the parentheses the understanding we detailed earlier stating that raising a fraction to the power of negative one will always give the reciprocal fraction, obtained by switching the numerator with the denominator,

Let's summarize the solution steps so far, we got that:

$\big(-\frac{4}{9}\big)^{-3}= (-1)^{-3}\cdot\big( \frac{4}{9}\big)^{-3} =-\big(\frac{9}{4}\big)^{3}$

Finally, let's recall the exponent law for exponents in parentheses with division of terms:

$\big(\frac{a}{c}\big)^n=\frac{a^n}{c^n}$

And let's apply this law to the expression we got in the last step:

$-\big(\frac{9}{4}\big)^3=-\frac{9^3}{4^3}$

Let's summarize the solution steps so far, we obtained the following:

$\big(-\frac{4}{9}\big)^{-3}=-\big(\frac{9}{4}\big)^{3}=-\frac{9^3}{4^3}$

Therefore the correct answer is answer C.