If b > 1, which of the following expressions are equivalent to \frac{\sqrt{b^3 \cdot \sqrt… — GRE General Test

If $b > 1$ , which of the following expressions are equivalent to $\frac{\sqrt{b^3 \cdot \sqrt[3]{b^2}}}{b^{-\frac{1}{6}}}$ ? Select all that apply.

The equivalent expressions are

(\sqrt[3]{b})^6

\frac{b^{\frac{7}{3}}}{b^{\frac{1}{3}}}

, and

\left(b^{-\frac{1}{2}}\right)^{-4}

Simplifying the original expression yields

\frac{\sqrt{b^3 \cdot b^{2/3}}}{b^{-1/6}} = \frac{b^{11/6}}{b^{-1/6}} = b^{11/6 - (-1/6)} = b^{12/6} = b^2

. The expression

(\sqrt[3]{b})^6

equals

b^{6/3} = b^2

. The expression

\frac{b^{7/3}}{b^{1/3}}

equals

b^{7/3 - 1/3} = b^2

. The expression

\left(b^{-1/2}\right)^{-4}

equals

b^{(-1/2)(-4)} = b^2

. All three equal

b^2

Simplify the expression inside the square root in the numerator.

b^3 \cdot \sqrt[3]{b^2} = b^3 \cdot b^{\frac{2}{3}} = b^{3 + \frac{2}{3}} = b^{\frac{11}{3}}

Convert radical to rational exponent and use product rule for exponents.

Apply the square root to the numerator.

\sqrt{b^{\frac{11}{3}}} = \left(b^{\frac{11}{3}}\right)^{\frac{1}{2}} = b^{\frac{11}{6}}

Taking the square root is equivalent to raising to the power of

\frac{1}{2}

Divide by the denominator.

\frac{b^{\frac{11}{6}}}{b^{-\frac{1}{6}}} = b^{\frac{11}{6} - \left(-\frac{1}{6}\right)} = b^{\frac{12}{6}} = b^2

Apply quotient rule for exponents: subtract the exponent of the denominator from the numerator.

Evaluate each given option to determine which simplify to

b^2

The expressions

(\sqrt[3]{b})^6 = b^2

\frac{b^{\frac{7}{3}}}{b^{\frac{1}{3}}} = b^2

, and

\left(b^{-\frac{1}{2}}\right)^{-4} = b^2

are all equivalent.

Matching each simplified candidate expression to the target simplified value

b^2

Simplification of nested algebraic radicals and rational exponents using exponent rules

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