Question

Which expression is equivalent to RootIndex 3 StartRoot StartFraction 75 a Superscript 7 Baseline b Superscript 4 Baseline Over 40 a Superscript 13 Baseline c Superscript 9 Baseline EndFraction EndRoot? Assume a not-equals 0 and c not-equals 0.
A. StartFraction a cubed b (RootIndex 3 StartRoot 15 b squared EndRoot) Over 2 c cubed EndFraction
B. StartFraction b (RootIndex 3 StartRoot 15 b EndRoot) Over 2 a squared c cubed EndFraction
C. StartFraction a cubed b (RootIndex 3 StartRoot 15 b squared EndRoot) Over 6 c cubed EndFraction
D. StartFraction b (RootIndex 3 StartRoot 15 b EndRoot) Over 2 a c EndFraction

Answer 1

Answer:

B. $\frac{b}{2a^{2}c^3}\sqrt[3]{15b}$

Step-by-step explanation:

Given:

The expression to simplify is given as:

$\sqrt[3]{\frac{75a^7b^4}{40a^{13}c^9}}$

Use the exponent property $\frac{a^m}{a^n}=a^{m-n}$

$\frac{a^7}{a^{13}}=a^{7-13}=a^{-6}$

Use the exponent property $(a^m)^n=a^{m\times n}$

$a^{-6}=a^{-2\times 3}=(a^{-2})^3$

$b^4=b\times b^3\\c^{9}=(c^3)^3$

Reducing $\frac{75}{40}$ to simplest form, we get:

$\frac{5\times 5\times 3}{2^3\times 5}=\frac{15}{2^3}$

Therefore, expression becomes:

$\sqrt[3]{\frac{15(a^{-2})^3\times b\times b^3}{2^3(c^3)^3}}$

Use the cubic root property:

$\sqrt[3]{x^3} =x$. Thus, the expression becomes:

$\frac{a^{-2}b}{2c^3}\sqrt[3]{15b}$

Using the exponent property $a^{-m}=\frac{1}{a^m}$

$a^{-2}=\frac{1}{a^2}$

So, the final expression is:

$\frac{b}{2a^{2}c^3}\sqrt[3]{15b}$

Therefore, the correct option is option B.