Question 44·Medium·Equivalent Expressions

Remember that $x^{\tfrac{2}{3}}$ means "take the cube root of $x$, then square it." Use this idea to simplify $27^{\tfrac{2}{3}}$, and multiply exponents for $a^{-3}$ and $b^{9}$.

After simplifying the numerator, write the whole fraction as one product in the numerator divided by $3b^{-1}$. For the $b$-terms, recall that when you divide powers with the same base, you subtract the exponent in the denominator from the exponent in the numerator.

Carefully simplify the numeric coefficient ($9$ divided by $3$) separately from the $a$ and $b$ exponents. Make sure you handle the negative exponent on $b^{-1}$ correctly when subtracting exponents.

In Desmos, pick specific nonzero values for $a$ and $b$ (for example, type a=2 and b=5). Then enter the original expression as ((27*a^(-3)*b^9)^(2/3))/(3*b^(-1)) and note its numeric value.

Still using the same $a$ and $b$, type each option into Desmos:

• 9*a^(-2)*b^7
• 3*a^2*b^5
• 3*a^(-2)*b^5
• 3*a^(-2)*b^7

Compare the value of each to the value of the original expression; the correct option will match it exactly.

Change $a$ and $b$ to a different pair of nonzero values (for example, a=-1, b=3) and check again. The expression that matches the original value for both sets of $(a,b)$ is the one that is algebraically equivalent.

Start by focusing on the numerator $\bigl(27a^{-3}b^{9}\bigr)^{\tfrac{2}{3}}$. Use the power rule $(xy)^n = x^n y^n$ to distribute the exponent $\tfrac{2}{3}$ to each factor:

Now simplify each factor separately:

• For $27^{\tfrac{2}{3}}$, remember that $x^{\tfrac{1}{3}}$ is the cube root of $x$:
  • The cube root of 27 is 3, so $27^{\tfrac{1}{3}} = 3$.
  • Then $27^{\tfrac{2}{3}} = (27^{\tfrac{1}{3}})^2 = 3^2 = 9$.
• For $(a^{-3})^{\tfrac{2}{3}}$, multiply the exponents: $(-3) \cdot \tfrac{2}{3} = -2$, so this becomes $a^{-2}$.
• For $(b^{9})^{\tfrac{2}{3}}$, multiply the exponents: $9 \cdot \tfrac{2}{3} = 6$, so this becomes $b^{6}$.

So the numerator simplifies to

Now place the simplified numerator over the original denominator $3b^{-1}$:

You can separate the constants, the $a$-terms, and the $b$-terms:

• Constants: $\dfrac{9}{3}$
• $a$-terms: $a^{-2}$ (there is no $a$ in the denominator)
• $b$-terms: $\dfrac{b^{6}}{b^{-1}}$

First simplify the constants:

• $\dfrac{9}{3} = 3$.

The $a$-term stays $a^{-2}$.

For the $b$-terms, use the rule $\dfrac{x^m}{x^n} = x^{m-n}$:

Putting everything together, the simplified expression is

which matches choice D.