(i) Yes \( \sqrt{3}+\sqrt{3} =2 \sqrt{3} =\sqrt{2^{2} \times 3} =\sqrt{12} \) pure sured
(ii) Yes \(2 \sqrt{5}-\sqrt{5}=\sqrt{5}\) pure sured
(iii) Yes \(\sqrt {5}\times\sqrt {7}=\sqrt {35}\), a pure sured
(iv) Yes, the quotient of two surds can be a pure surd. \(\frac{\sqrt{5}}{\sqrt{2}}=\sqrt{\frac{5}{2}}\)

(i) \(\sqrt [ 3 ]{ 5 } ,\sqrt [ 9 ]{ 4 } ,\sqrt [ 6 ]{ 3 } \)
\(5^{\frac{1}{3}}\)
ஃ The order of the surds \(\sqrt [ 3 ]{ 5 } ,\sqrt [ 9 ]{ 4 } ,\sqrt [ 6 ]{ 3 } \) are 3,9,6
\(4^{\frac{1}{9}}\)
\(3^{\frac{1}{6}}\) l.c.m of 3,9,6 is 18
ஃ \(\frac{1}{3}=\frac{1\times6}{3\times6}=\frac{6}{18}\)
\(\frac{1}{9}=\frac{1\times2}{9\times2}=\frac{2}{18};\frac{1}{6}=\frac{1\times3}{6\times3}=\frac{3}{18}\)
\((5^{\frac{1}{3}})=5^{\frac{6}{18}}=(15625)^{\frac{1}{18}}\)
\((4^{\frac{1}{9}})=4^{\frac{2}{18}}=(4^2)^{\frac{1}{18}}=16^{\frac{1}{18}}\)
\((3^{\frac{1}{6}})=3^{\frac{3}{18}}=(3^3)^{\frac{1}{18}}=27^{\frac{1}{18}}\)
ஃ The descending order of \(\sqrt [ 3 ]{ 5 } ,\sqrt [ 9 ]{ 4 } ,\sqrt [ 6 ]{ 3 } \) is \((15625)^{\frac{1}{18}}>(27)^{\frac{1}{18}}>16^{\frac{1}{18}}\) i.e., \(\sqrt { \sqrt { 3 } } >\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } >\sqrt [ 3 ]{ \sqrt [ 4 ]{ 7 } } \)
(ii) \(\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 3 ]{ \sqrt [ 4 ]{ 7 } } ,\sqrt { \sqrt { 3 } } \)
The order of the surds \(\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 3 ]{ \sqrt [ 4 ]{ 7 } } ,\sqrt[2] { \sqrt [2]{ 3 } } \)  are 6, 12, 4
l.c.m of 6,12,4 is 12
\(\sqrt[2]{\sqrt[3]{5}}=5^{\frac{1}{6}}=5^{\frac{1\times2}{6\times2}}=5^{\frac{2}{12}}=(5^2)^{\frac{1}{12}}=25^{\frac{1}{12}}\)
\(\sqrt[3]{\sqrt[4]{\sqrt{7}}}=7^{\frac{1}{12}};\sqrt{\sqrt{3}}=3^{\frac{1}{4}}=3^{\frac{1\times3}{4\times3}}=3^{\frac{3}{12}}=(3^3)^{\frac{1}{12}}=27^{\frac{1}{12}}\)
ஃ The ascending order of the surds
\(\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 2 ]{ \sqrt [ 3 ]{ 5 } } ,\sqrt [ 3 ]{ \sqrt [ 4 ]{ 7 } } ,\sqrt { \sqrt { 3 } } \) is \(7^{\frac{1}{12}}<25^{\frac{1}{2}}<27^{\frac{1}{2}}\), that is \(\sqrt[3]{\sqrt[4]{7}}<\sqrt[2]{\sqrt[3]{5}}<\sqrt{\sqrt{3}}\)

(i) \(\sqrt{3}\times\sqrt{5}\times\sqrt{2}=\sqrt{3\times5\times2}=\sqrt{30}\)
(ii) \(\sqrt{35}\div\sqrt{7}=\sqrt{\frac{35}{7}}=\sqrt{5}\)
(iii) \(\sqrt[3]{27}\times\sqrt[3]{8}\times\sqrt[3]{125}=\sqrt[3]{27\times8\times125}=\sqrt[3]{3^3\times2^3\times5^3}=3\times2\times5=30\)
(iv) \((7\sqrt{a}-5\sqrt{b})(7\sqrt{a}+5\sqrt{b})=(7\sqrt{a})^2-(5\sqrt{b})^2=49a-25b\) = 21\(\sqrt 3\)
(v) \([\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}]\div\sqrt{\frac{16}{81}}\)

=\([\sqrt{\frac{15^2}{27^2}}-\sqrt{\frac{5^2}{12^2}}]\times\sqrt{\frac{9^2}{4^2}}\)
=\((\frac{15}{27}-\frac{5}{12})\times\frac{9}{4}=(\frac{5}{9}-\frac{5}{12})\times\frac{9}{4}\)
=\((\frac{20-15}{36})\times\frac{9}{4}=\frac{5}{ ̶3̶6̶}\times\frac{ ̶9̶}{4}=\frac{5}{16}\)

(i) (5 + 18 - 2)√3 = 21\(\sqrt{3}\)
(ii) (4 + 2 - 3)\(\sqrt[3]{5}\) = \(3\sqrt [ 3 ]{ 5 } \)

(iii) =\(3\sqrt{5\times5\times3}+5\sqrt{3\times2\times2\times2\times2}-\sqrt{3\times3\times3\times3\times3}\)
= 3\(\times\)5√3 + 5\(\times\)2\(\times\)2√3 - 3\(\times\)3√3
= 15√3 + 20√3 - 9√3
= (15 + 20 - 9)√3 = 26\(\sqrt{3}\)
(iv) \(5\sqrt[3]{2^3\times5}+2\sqrt[3]{5^3\times5}-3\sqrt[3]{2^3\times2^3\times5}\)
= \(5\times2\times\sqrt[3]{5}+2\times5\sqrt[3]{5}-3\times2\times2\sqrt[3]{5}\) = \(8\sqrt [ 3 ]{ 5 } \)

\(\sqrt[5]{\frac{1024}{3125}}=(\frac{1024}{3125})^{\frac{1}{5}}=((\frac{4}{5})^{ ̶5̶})^{\frac{1}{ ̶5̶}}=\frac{4}{5}\)

\(\sqrt[5]{100000}=(100000)^{\frac{1}{5}}=(10^{ ̶5̶})^{\frac{1}{ ̶5̶}}=10\)

\(\sqrt[5]{243}=243^{\frac{1}{5}}=(3^5)^{\frac{1}{5}}=3^{ ̶5̶\times\frac{1}{ ̶5̶}}=3\)

\(\sqrt[5]{32}=32^{\frac{1}{5}}=(2^5)^{\frac{1}{5}}=2^{ ̶5̶\times\frac{1}{ ̶5̶}}=2\)