Question 12·Hard·Equivalent Expressions

Remember: when dividing like bases, subtract exponents (numerator minus denominator), and when you raise a power to another power, multiply the exponents. Keep track of negative signs in each step.

After you simplify inside, apply the $-3$ exponent to the constant, the $a$ part, and the $b$ part separately. Then think about how to rewrite any negative exponents so that all exponents are positive.

Once you have an expression with fractional exponents, rewrite exponents like $m + 1/4$ or $m + 1/2$ as a product of an integer power and a radical to compare directly with the answer choices.

Choose convenient positive numbers, such as $a=2$ and $b=3$ (any positive values work as long as they are used consistently).

In Desmos, type ( (2*a^(-1/4)*b^(2/3)) / (8*a^(5/6)*b^(-1/2)) )^(-3) and then, on another line, assign a=2 and b=3 to get a numerical value for the original expression.

On separate lines, enter each choice using the same values for a and b, for example: 64*a^3*sqrt(a)/ (b^3*sqrt[4](b)) for choice A (use Desmos’s sqrt() and ^(1/4) for fourth roots if needed).

Compare the numerical value of the original expression with the values of the four answer choices. The option whose value matches the original expression’s value (for your chosen $a$ and $b$) is the correct one.

Start with $\left( \dfrac{2a^{-1/4} b^{2/3}}{8 a^{5/6} b^{-1/2}} \right)^{-3}$. First simplify the constant and exponents inside the parentheses:

• The constant: $2/8 = 1/4$.
• For $a$: when dividing, subtract exponents: $-\tfrac{1}{4} - \tfrac{5}{6} = -\tfrac{3}{12} - \tfrac{10}{12} = -\tfrac{13}{12}$.
• For $b$: $\tfrac{2}{3} - ( -\tfrac{1}{2}) = \tfrac{2}{3} + \tfrac{1}{2} = \tfrac{4}{6} + \tfrac{3}{6} = \tfrac{7}{6}$. So the inside becomes $(\tfrac{1}{4}) a^{-13/12} b^{7/6}$, all raised to the power $-3$.

Use the rule $(xy)^n = x^n y^n$ and $(x^m)^n = x^{mn}$.

• For the constant: $(\tfrac{1}{4})^{-3} = 4^3 = 64$.
• For $a$: $(a^{-13/12})^{-3} = a^{(-13/12)(-3)} = a^{39/12} = a^{13/4}$.
• For $b$: $(b^{7/6})^{-3} = b^{(7/6)(-3)} = b^{-21/6} = b^{-7/2}$. So the expression becomes $64 a^{13/4} b^{-7/2}$.

A negative exponent means the base moves to the denominator: $x^{-k} = 1/x^k$. Here $b^{-7/2}$ moves to the denominator as $b^{7/2}$, so the expression is $\dfrac{64 a^{13/4}}{b^{7/2}}$. Now we just need to convert the fractional exponents $13/4$ and $7/2$ into a product of an integer power and a root to match the answer choices.

Break each fractional exponent into an integer plus a fraction:

• $a^{13/4} = a^{3 + 1/4} = a^3 a^{1/4} = a^3 \sqrt[4]{a}$.
• $b^{7/2} = b^{3 + 1/2} = b^3 b^{1/2} = b^3 \sqrt{b}$. Substitute these into $\dfrac{64 a^{13/4}}{b^{7/2}}$ to get $\dfrac{64 a^3 \sqrt[4]{a}}{b^3 \sqrt{b}}$, which matches choice D.