Question

$\sqrt[3]{3\frac{-a^6\times b^3\times c^{21}}{c^9\times a^{12}}}=$

• A. $\frac{-b{c}^3}{a^2}$
• B. $\frac{b{c}^4}{a^2}$
• C. $\frac{-a{b}^4}{c^2}$
• D. $\frac{-b{c}^4}{a^2}$

Answer 1

Answer: (D)

$\sqrt[3]{3\frac{-a^6\times b^3\times c^{21}}{c^9\times a^{12}}}$

$=\sqrt{-a^{6-12}\times b^3\times c^{21-9}}$ [Since, $\frac{x^m}{x^n}=x^{m-n},x\ne 0$]

$=\sqrt[3]{3-a^{-6}\times b^3\times c^{12}}$

$=\sqrt[3]{3\frac{b^3\times c^{12}}{-a^6}}$ [Since,$a^{-m}=\frac{1}{a^m}$]

$=(\frac{b^3\times c^{12}}{-a^6}{)}^{\frac{1}{3}}$

$=\frac{b^{3\times \frac{1}{3}}\times c^{12\times \frac{1}{3}}}{-a^{6\times \frac{1}{3}}}$ [Since, $(a^m{)}^n=a^{m\times n}$]

$=-\frac{b{c}^4}{a^2}$

$\therefore \sqrt[3]{3\frac{-a^6\times b^3\times c^{21}}{c^9\times a^{12}}}=-\frac{b{c}^4}{a^2}$

Hence, Option D is correct.