Solve the Mixed Fraction and Fourth Root Equation: 2^3/3^2 * 3^-2 * Fourth Root of 81 = ?

First, let's note that in the problem there are terms with bases 2 and 3, and a term with base 81 which is inside the root,

Next, let's note that the number 81 is a power of the number 3:

$81=3^4$

Therefore we can replace it with this power of 3 in order to get a term with base 3, let's apply this in the problem:

$\frac{2^3}{3^2}\cdot3^{-2}\cdot\sqrt[4]{81}=\frac{2^3}{3^2}\cdot3^{-2}\cdot\sqrt[4]{3^4}$

Next we want to get rid of the root in the problem so we'll recall the definition of the nth root as a power:

$\sqrt[n]{a}=a^{\frac{1}{n}}$

Let's apply this law to the third term from the left in the product in the expression we got in the last step:

$\frac{2^3}{3^2}\cdot3^{-2}\cdot\sqrt[4]{3^4}=\frac{2^3}{3^2}\cdot3^{-2}\cdot \big(3^4\big)^{\frac{1}{4}}$

When we did this carefully since the term inside the root is a term with a power, therefore we used parentheses,

Next let's recall the power law for power of a power:

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

And let's apply this law in the expression we got in the last step on the same term in the product we dealt with until now:

$\frac{2^3}{3^2}\cdot3^{-2}\cdot\big(3^4\big)^{\frac{1}{4}}=\frac{2^3}{3^2}\cdot3^{-2}\cdot3^{4\cdot\frac{1}{4}}=\frac{2^3}{3^2}\cdot3^{-2}\cdot3^{\frac{4}{4}}=\frac{2^3}{3^2}\cdot3^{-2}\cdot3^1=\frac{2^3}{3^2}\cdot3^{-2}\cdot3$

When in the first stage we applied the above law on the third term in the product and then simplified the expression in the power exponent of that term while remembering that multiplication in a fraction means multiplication in the fraction's numerator,

Now let's recall the power law for negative power:

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

And we'll represent using it the last two terms in the product in the expression we got in the last step, as fractions, in order to later perform fraction multiplication:

$\frac{2^3}{3^2}\cdot3^{-2}\cdot3=\frac{2^3}{3^2}\cdot\frac{1}{3^2}\cdot\frac{1}{3^{-1}}$

When for the second term from the left in the product we applied the above power law directly, and in the third term we applied it while understanding that we can represent the number 3 as a term with a negative power in the following way:

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

When we performed the use of the negative power law carefully, since the number that n represents in the above power law in our use here is:

$-1$

(Marked with underline in the expression above)

Let's summarize the solution steps until here, we got that:

$\frac{2^3}{3^2}\cdot3^{-2}\cdot\sqrt[4]{3^4}=\frac{2^3}{3^2}\cdot3^{-2}\cdot \big(3^4\big)^{\frac{1}{4}} =\frac{2^3}{3^2}\cdot3^{-2}\cdot3 = \frac{2^3}{3^2}\cdot\frac{1}{3^2}\cdot\frac{1}{3^{-1}}$

Next we'll perform the fraction multiplication while remembering the rule for multiplying fractions:

$\frac{a}{b}\cdot\frac{c}{d}=\frac{a\cdot c}{b\cdot d}$

Let's apply this rule in the last expression we got:

$\frac{2^3}{3^2}\cdot\frac{1}{3^2}\cdot\frac{1}{3^{-1}}=\frac{2^3\cdot1\cdot1}{3^2\cdot3^2\cdot3^{-1}}=\frac{2^3}{3^2\cdot3^2\cdot3^{-1}}$

Next let's recall the power law for multiplication between terms with equal bases:

$a^m\cdot a^n=a^{m+n}$

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

$\frac{2^3}{3^2\cdot3^2\cdot3^{-1}}=\frac{2^3}{3^{2+2+(-1)}}=\frac{2^3}{3^{2+2-1}}=\frac{2^3}{3^3}$$$

When in the first stage we applied the above power law in the fraction's denominator and in the following stages we simplified the resulting expression,

Finally let's recall another important power law - it's the power law for power applied to parentheses:

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

And let's apply it in the expression we got in the last step:

$\frac{2^3}{3^3}=\big(\frac{2}{3}\big)^3$

When we applied the above power law in the opposite direction, meaning - instead of opening the parentheses and applying the power to the fraction's numerator and denominator, we used the fact that both the numerator and denominator are raised to the same power and therefore we can represent the expression as a fraction raised to a power, which can be done only because both the numerator and denominator are raised to the same power.

Let's summarize the solution steps until here, we got that:

$\frac{2^3}{3^2}\cdot3^{-2}\cdot\sqrt[4]{3^4}=\frac{2^3}{3^2}\cdot3^{-2}\cdot3 = \frac{2^3}{3^2\cdot3^2\cdot3^{-1}} =\frac{2^3}{3^3} = \big(\frac{2}{3}\big)^3$

Therefore the correct answer is answer A.