Question

13.......................

Answer 1

Answer:

The answer is (d) ⇒ $pq^{2}r\sqrt[3]{pr^{2}}$

Step-by-step explanation:

* To simplify the cube roots:

 If its number then the number must be written in the form x³

 then we divide the power by 3 to cancel the radical

 If its variable we divide its power by 3 to cancel the radical

∵ $\sqrt[3]{p^{4}q^{6}r^{5}}=p^{\frac{4}{3}}q^{\frac{6}{3}}r^{\frac{5}{3}}}$

∴ $p^{\frac{4}{3}}q^{2}}r^{\frac{5}{3}}=p^{1\frac{1}{3}}q^{2}r^{1\frac{2}{3}}$

∵ $p^{\frac{1}{3}}=\sqrt[3]{p}$

∵ $r^{\frac{2}{3}}=\sqrt[3]{r^{2}}$

∴ $p(p)^{\frac{1}{3}}q^{2}r(r)^{\frac{2}{3}}=p(\sqrt[3]{p})q^{2}r(\sqrt[3]{r^{2}})$

∴ $prq^{2}\sqrt[3]{pr^{2}}}$

∴ The answer is (d)