Solution:
$$\eqalign{ \root 4 \of {3x + 1} = 2 \cr \Rightarrow {\left( {\root 4 \of {3x + 1} } \right)^4} = {2^4} \cr \Rightarrow {\left( {3x + 1} \right)^{4 \times \frac{1}{4}}} = 16 \cr \Rightarrow 3x = 15 \cr \Rightarrow x = 5 \cr} $$

Solution:
Given expression,
$${\frac{1}{{1 + a + {b^{ - 1}}}} + }$$   $${\frac{1}{{1 + b + {c^{ - 1}}}} + }$$   $${\frac{1}{{1 + c + {a^{ - 1}}}}}$$
  $$ = \frac{1}{{1 + a + {b^{ - 1}}}} + $$   $$\frac{b^{ - 1}}{{{b^{ - 1}} + 1 + {b^{ - 1}}{c^{ - 1}}}} + $$    $$\frac{1}{{a + ac + 1}}$$
  $$ = \frac{1}{{1 + a + {b^{ - 1}}}} + $$   $$\frac{{{b^{ - 1}}}}{{1 + {b^{ - 1}} + a}} + $$   $$\frac{a}{{a + {b^{ - 1}} + 1}}$$
$$\eqalign{ = \frac{{1 + a + {b^{ - 1}}}}{{1 + a + {b^{ - 1}}}} \cr = 1 \cr} $$
$$\left[ {\because abc = 1 \Rightarrow {{\left( {bc} \right)}^{ - 1}} = a \Rightarrow {b^{ - 1}}{c^{ - 1}} = a,{\text{and }}ac = {b^{ - 1}}} \right]$$

Solution:
$$\eqalign{ \frac{{256 \times 256 - 144 \times 144}}{{112}} \cr = \frac{{{{\left( {256} \right)}^2} - {{\left( {144} \right)}^2}}}{{112}} \cr = \frac{{\left( {112} \right)\left( {400} \right)}}{{112}} \cr = 400 \cr} $$

Solution:
$$\eqalign{ {\text{Let }}{8^{2.4}} \times {2^{3.7}} \div {\left( {16} \right)^{1.3}} = {2^x} \cr {\text{Then,}}{\left( {{2^3}} \right)^{2.4}} \times {2^{3.7}} \div {\left( {{2^4}} \right)^{1.3}} = {2^x} \cr \Leftrightarrow {2^{\left( {3 \times 2.4} \right)}} \times {2^{3.7}} \div {2^{\left( {4 \times 1.3} \right)}} = {2^x} \cr \Leftrightarrow {2^{7.2}} \times {2^{3.7}} \div {2^{5.2}} = {2^x} \cr \Leftrightarrow {2^x} = {2^{\left( {7.2 + 3.7 - 5.2} \right)}} \cr \Leftrightarrow {2^x} = {2^{5.7}} \cr \Leftrightarrow x = 5.7 \cr} $$

Solution:
$$\eqalign{ a = \frac{1}{{2 - \sqrt 3 }} + \frac{1}{{3 - \sqrt 8 }} + \frac{1}{{4 - \sqrt {15} }} \cr \Rightarrow \frac{1}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} + \frac{1}{{3 - \sqrt 8 }} \times \frac{{3 + \sqrt 8 }}{{3 + \sqrt 8 }} + \frac{1}{{4 - \sqrt {15} }} \times \frac{{4 + \sqrt {15} }}{{4 + \sqrt {15} }} \cr \Rightarrow \frac{{2 + \sqrt 3 }}{{4 - 3}} + \frac{{3 + \sqrt 8 }}{{9 - 8}} + \frac{{4 + \sqrt {15} }}{{16 - 15}} \cr \Rightarrow 2 + \sqrt 3 + 3 + \sqrt 8 + 4 + \sqrt {15} \cr \Rightarrow 9 + \sqrt 3 + 2\sqrt 2 + \sqrt {15} \cr a = 9 < 9 + \sqrt 3 + 2\sqrt 2 + \sqrt {15} < 18 \cr \sqrt 3 = 1.73,\,\sqrt 2 = 1.41,\,\sqrt {15} = 3.9 \cr \Rightarrow 9 < 9 + 1.73 + \left( {2 \times 1.41} \right) + 3.9 = 17.4 < 18 \cr} $$

Solution:
$$\eqalign{ {\text{I}}.\sqrt {12} > \root 3 \of {16} > \root 4 \of {24} \cr {12^{\frac{1}{2}}},\,{16^{\frac{1}{3}}},\,{24^{\frac{1}{4}}} \cr 2{\left( 3 \right)^{\frac{1}{2}}} > 2{\left( 2 \right)^{\frac{1}{3}}} > 2{\left( {\frac{3}{2}} \right)^{\frac{1}{4}}}\,{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {\text{I}} \right){\text{ True}} \cr {\text{II}}.\root 3 \of {25} > \root 4 \of {32} > \root 6 \of {48} \cr {25^{\frac{1}{3}}},\,{32^{\frac{1}{4}}},\,{48^{\frac{1}{6}}} \cr 5{\left( {\frac{1}{5}} \right)^{\frac{1}{3}}} > 2{\left( 2 \right)^{\frac{1}{4}}} > 2{\left( {\frac{3}{4}} \right)^6}\,{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{II}}} \right){\text{ True}} \cr {\text{III}}.\root 4 \of 9 > \root 3 \of {15} > \root 6 \of {24} \cr {9^{\frac{1}{4}}},\,{15^{\frac{1}{3}}},\,{24^{\frac{1}{6}}} \cr {\text{3}}{\left( {\frac{1}{9}} \right)^{\frac{1}{4}}} > {\left( {15} \right)^{\frac{1}{3}}} > {\left( {24} \right)^{\frac{1}{6}}}\,{\text{. }}{\text{. }}{\text{. }}{\text{. }}\left( {{\text{III}}} \right){\text{ False}} \cr} $$

Solution:
$$\eqalign{ \left( {\sqrt 2 + \sqrt 5 - \sqrt 3 } \right) \times {\text{k}} = - 12 \cr {\text{In these type of questions, we can go through option}} \cr {\text{From option D}} \cr \Rightarrow \left( {\sqrt 2 + \sqrt 5 - \sqrt 3 } \right)\left( {\sqrt 2 + \sqrt 5 + \sqrt 3 } \right)\left( {2 - \sqrt {10} } \right) = - 12 \cr \Rightarrow \left[ {{{\left( {\sqrt 2 + \sqrt 5 } \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} \right]\left( {2 - \sqrt {10} } \right) = - 12 \cr \Rightarrow \left( {2 + 5 + 2\sqrt {10} - 3} \right)\left( {2 - \sqrt {10} } \right) = - 12 \cr \Rightarrow 2\left( {2 + \sqrt {10} } \right)\left( {2 - \sqrt {10} } \right) = - 12 \cr \Rightarrow 2\left[ {{{\left( 2 \right)}^2} - {{\left( {\sqrt {10} } \right)}^2}} \right] = - 12 \cr \Rightarrow 2 \times \left[ { - 6} \right] = - 12 \cr \Rightarrow - 12 = - 12\,\,\,\,\left( {{\text{Satisfy}}} \right) \cr} $$