$$\eqalign{ & \frac{{{2^{n - 1}} - {2^n}}}{{{2^{n + 4}} + {2^{n + 1}}}} \cr & = \frac{{{2^{n - 1}}\left( {1 - 2} \right)}}{{{2^{n + 1}}\left( {{2^3} + 1} \right)}} \cr & = \left( { - \frac{1}{9}} \right){.2^{\left( {n - 1} \right) - \left( {n + 1} \right)}} \cr & = \left( { - \frac{1}{9}} \right){.2^{ - 2}} \cr & = \left( { - \frac{1}{9}} \right).\frac{1}{{{2^2}}} \cr & = \left( { - \frac{1}{9}} \right) \times \frac{1}{4} \cr & = - \frac{1}{{36}} \cr} $$

$$\eqalign{ & {\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2} \cr & = x + \frac{1}{x} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \frac{1}{{\left( {5 + 2\sqrt 6 } \right)}} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \frac{1}{{\left( {5 + 2\sqrt 6 } \right)}} \times \frac{{\left( {5 - 2\sqrt 6 } \right)}}{{\left( {5 - 2\sqrt 6 } \right)}} - 2 \cr & = \left( {5 + 2\sqrt 6 } \right) + \left( {5 - 2\sqrt 6 } \right) - 2 \cr & = 10 - 2 \cr & = 8 \cr & \therefore \left( {\sqrt x - \frac{1}{{\sqrt x }}} \right) = \sqrt 8 = 2\sqrt 2 \cr} $$

$$\eqalign{ & {\text{Let }}{\left( {0.04} \right)^2} \div \left( {0.008} \right) \times {\left( {0.2} \right)^6} = {\left( {0.2} \right)^x} \cr & {\text{Then,}}{\left( {0.2} \right)^x} = {\left[ {{{\left( {0.2} \right)}^2}} \right]^2} \div {\left( {0.2} \right)^3} \times {\left( {0.2} \right)^6} \cr & \Leftrightarrow {\left( {0.2} \right)^x} = {\left( {0.2} \right)^{\left( {2 \times 2} \right)}} \div {\left( {0.2} \right)^3} \times {\left( {0.2} \right)^6} \cr & \Leftrightarrow {\left( {0.2} \right)^x} = {\left( {0.2} \right)^4} \div {\left( {0.2} \right)^3} \times {\left( {0.2} \right)^6} \cr & \Leftrightarrow {\left( {0.2} \right)^x} = {\left( {0.2} \right)^{\left( {4 - 3 + 6} \right)}} \cr & \Leftrightarrow {\left( {0.2} \right)^x} = {\left( {0.2} \right)^7} \cr & \Leftrightarrow x = 7 \cr} $$

$$\eqalign{ & \frac{{{{\left( {243} \right)}^{0.13}} \times {{\left( {243} \right)}^{0.07}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( {49} \right)}^{0.075}} \times {{\left( {343} \right)}^{0.2}}}} \cr & = \frac{{{{\left( {243} \right)}^{\left( {0.13 + 0.07} \right)}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( {{7^2}} \right)}^{0.075}} \times {{\left( {{7^3}} \right)}^{0.2}}}} \cr & = \frac{{{{\left( {243} \right)}^{0.2}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( 7 \right)}^{\left( {2 \times 0.075} \right)}} \times {{\left( 7 \right)}^{\left( {3 \times 0.2} \right)}}}} \cr & = \frac{{{{\left( {{3^5}} \right)}^{0.02}}}}{{{{\left( 7 \right)}^{0.25}} \times {{\left( 7 \right)}^{0.15}} \times {{\left( 7 \right)}^{0.6}}}} \cr & = \frac{{{{\left( 3 \right)}^{\left( {5 \times 0.2} \right)}}}}{{{{\left( 7 \right)}^{\left( {0.25 + 0.15 + 0.6} \right)}}}} \cr & = \frac{{{3^1}}}{{{7^1}}} \cr & = \frac{3}{7} \cr} $$

$$\eqalign{ & {{\text{3}}^{x - y}} = 27 = {3^3} \cr & \Leftrightarrow x - y = 3........(i) \cr & {3^{x + y}} = 243 = {3^5} \cr & \Leftrightarrow x + y = 5........(ii) \cr & {\text{On solving (i) and (ii) ,}} \cr & {\text{we get }}x = 4 \cr} $$

$$\eqalign{ & {a^1} \cr & = {c^z} \cr & = {\left( {{b^y}} \right)^z} \cr & = {b^{yz}} \cr & = {\left( {{a^x}} \right)^{yz}} \cr & = {a^{xyz}} \cr & \Rightarrow xyz = 1 \cr} $$

$$\eqalign{ & \left( {\sqrt 3 + 1} \right)\left( {10 + \sqrt {12} } \right)\left( {\sqrt {12} - 2} \right)\left( {5 - \sqrt 3 } \right) \cr & \Rightarrow \left( {\sqrt 3 + 1} \right)\left( {10 + 2\sqrt 3 } \right)\left( {2\sqrt 3 - 2} \right)\left( {5 - \sqrt 3 } \right) \cr} $$
$$ \Rightarrow \left( {\sqrt 3 + 1} \right) \times $$   $$2\left( {5 + \sqrt 3 } \right) \times $$   $$2\left( {\sqrt 3 - 1} \right)$$  $$\left( {5 - \sqrt 3 } \right)$$
$$\eqalign{ & \Rightarrow 4\left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 - 1} \right)\left( {5 + \sqrt 3 } \right)\left( {5 - \sqrt 3 } \right) \cr & \Rightarrow 4\left[ {{{\left( {\sqrt 3 } \right)}^2} - {1^2}} \right]\left[ {{{\left( 5 \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}} \right] \cr & \Rightarrow 4 \times 2 \times 22 \cr & \Rightarrow 176 \cr} $$

$$\eqalign{ & \frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} \cr & = \frac{1}{{1 + \frac{{{a^n}}}{{{a^m}}}}} + \frac{1}{{1 + \frac{{{a^m}}}{{{a^n}}}}} \cr & = \frac{{{a^m}}}{{{a^m} + {a^n}}} + \frac{{{a^n}}}{{{a^m} + {a^n}}} \cr & = \frac{{\left( {{a^m} + {a^n}} \right)}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = 1 \cr} $$

$$\eqalign{ & \frac{{4 + 3\sqrt 3 }}{{\sqrt {7 + 4\sqrt 3 } }}{\text{ = }}A + \sqrt B \cr & \Rightarrow \sqrt {7 + 4\sqrt 3 } \cr & \Rightarrow \sqrt {{2^2} + {{\left( {\sqrt 3 } \right)}^2} + 2 \times 2\sqrt 3 } \cr & \Rightarrow \sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} \cr & \Rightarrow \left( {2 + \sqrt 3 } \right) \cr & \Rightarrow \frac{{4 + 3\sqrt 3 }}{{2 + \sqrt 3 }}{\text{ = }}A + \sqrt B \cr & \Rightarrow \frac{{4 + 3\sqrt 3 }}{{2 + \sqrt 3 }} \times \frac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }} = A + \sqrt B \cr & \Rightarrow \frac{{\left( {4 + 3\sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)}}{{4 - 3}} = A + \sqrt B \cr & \Rightarrow 8 - 4\sqrt 3 + 6\sqrt 3 - 9 = A + \sqrt B \cr & \Rightarrow 2\sqrt 3 - 1 = A + \sqrt B \cr & \therefore A = - 1\,\,\& \,\,\sqrt B = 2\sqrt 3 \cr & \because B = 2\sqrt 3 \times 2\sqrt 3 = 12 \cr & {\text{So}},B - A = 12 - \left( { - 1} \right) = 13 \cr} $$

$$\eqalign{ & {x^z}{\text{ = }}{y^2}{\text{ }} \cr & \Leftrightarrow {\left( {{{10}^{0.48}}} \right)^z} = {\left( {{{10}^{0.70}}} \right)^2} \cr & \Leftrightarrow {10^{\left( {0.48z} \right)}} = {10^{\left( {2 \times 0.70} \right)}} = {10^{1.40}} \cr & \Leftrightarrow 0.48z = 1.40 \cr & \Leftrightarrow z = \frac{{140}}{{48}} = \frac{{35}}{{12}} \cr & \Leftrightarrow z = 2.9\left( {{\text{approx}}} \right) \cr} $$

$$\eqalign{ & \sqrt {6 - 4\sqrt 3 + \sqrt {16 - 8\sqrt 3 } } \cr & = \sqrt {6 - 4\sqrt 3 + \sqrt {12 + 4 - 8\sqrt 3 } } \cr & = \sqrt {6 - 4\sqrt 3 + \sqrt {{{\left( {2\sqrt 3 } \right)}^2} + {{\left( 2 \right)}^2} - 2 \times 2\sqrt 3 \times 2} } \cr & = \sqrt {6 - 4\sqrt 3 + \sqrt {{{\left( {2\sqrt 3 - 2} \right)}^2}} } \cr & = \sqrt {6 - 4\sqrt 3 + 2\sqrt 3 - 2} \cr & = \sqrt {{{\left( {\sqrt 3 } \right)}^2} + {{\left( 1 \right)}^2} - 2 \times \sqrt 3 \times 1} \cr & = \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} \cr & = \sqrt 3 - 1 \cr} $$

$$\eqalign{ & {\text{Let X = }}\frac{{\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} }}{{\sqrt {\sqrt 5 + 1} }} \cr & {\text{Then,}} \cr & {{\text{X}}^2} = \frac{{{{\left( {\sqrt {\sqrt 5 + 2} + \sqrt {\sqrt 5 - 2} } \right)}^2}}}{{{{\left( {\sqrt {\sqrt 5 + 1} } \right)}^2}}} \cr} $$
  $${{\text{X}}^2} = {\text{ }}\frac{{\left( {\sqrt 5 + 2} \right) + \left( {\sqrt 5 - 2} \right) + 2\sqrt {\left( {\sqrt 5 + 2} \right)\left( {\sqrt 5 - 2} \right)} }}{{\left( {\sqrt 5 + 1} \right)}}$$
$$\eqalign{ & {{\text{X}}^2} = \frac{{2\sqrt 5 + 2\sqrt {{{\left( {\sqrt 5 } \right)}^2} - {{\left( 2 \right)}^2}} }}{{\sqrt 5 + 1}} \cr & {{\text{X}}^2} = \frac{{2\sqrt 5 + 2}}{{\sqrt 5 + 1}} \cr & {{\text{X}}^2} = \frac{{2\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 + 1} \right)}} \cr & {{\text{X}}^2} = 2 \cr & {\text{X = }}\sqrt 2 \cr & \therefore {\text{N}} = \sqrt 2 - \sqrt {3 - 2\sqrt 2 } \cr & {\text{N}} = \sqrt 2 - \sqrt {{{\left( {\sqrt 2 } \right)}^2} + {1^2} - 2 \times \sqrt 2 \times } 1 \cr & {\text{N}} = \sqrt 2 - \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} \cr & {\text{N}} = \sqrt 2 - \left( {\sqrt 2 - 1} \right) \cr & {\text{N}} = 1 \cr} $$

$$\eqalign{ & \left( {4 + \sqrt 7 } \right) \cr & = \frac{7}{2} + \frac{1}{2} + 2 \times \frac{{\sqrt 7 }}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \cr & = {\left( {\frac{{\sqrt 7 }}{{\sqrt 2 }}} \right)^2} + {\left( {\frac{1}{{\sqrt 2 }}} \right)^2} + 2 \times \frac{{\sqrt 7 }}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} \cr & = {\left( {\frac{{\sqrt 7 }}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}} \right)^2} \cr & = \frac{1}{2}{\left( {\sqrt 7 + 1} \right)^2} \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \Rightarrow {{\text{3}}^{x + y}}{\text{ = 81}}\,{\text{and}}\,{\text{8}}{{\text{1}}^{x - y}}{\text{ = 3}} \cr & \Rightarrow {{\text{3}}^{x + y}}{\text{ = (3}}{{\text{)}}^4}\,{\text{and}}\,{\left( 3 \right)^{4(}}^{x - y)}{\text{ = }}{{\text{3}}^1} \cr & \Rightarrow x + y = 4\,{\text{and}}\,x - y = \frac{1}{4} \cr & x + y = 4......{\text{(i)}} \cr & {\text{ }}x - y = \frac{1}{4}.....(ii) \cr & {\text{Solve the equation of (i) and (ii)}} \cr & x = \frac{{17}}{8}, \cr & y = \frac{{15}}{8}, \cr & \Rightarrow \frac{x}{y} = \frac{{17}}{{15}} \cr} $$

$$\eqalign{ & \frac{{{2^{n + 4}} - 2 \times {2^n}}}{{2 \times {2^{\left( {n + 3} \right)}}}} + {2^{ - 3}} \cr & = \frac{{{2^{n + 4}} - {2^{n + 1}}}}{{{2^{\left( {n + 4} \right)}}}} + \frac{1}{{{2^3}}} \cr & = \frac{{{2^{n + 1}}\left( {{2^3} - 1} \right)}}{{{2^{\left( {n + 4} \right)}}}} + \frac{1}{{{2^3}}} \cr & = \frac{{{2^{n + 1}} \times 7}}{{{2^{n + 1}} \times {2^3}}} + \frac{1}{{{2^3}}} \cr & = \left( {\frac{7}{8} + \frac{1}{8}} \right) \cr & = \frac{8}{8} \cr & = 1 \cr} $$

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

$$\eqalign{ & {\text{Given expression, }} \cr & \frac{1}{{1 + \frac{{{x^b}}}{{{x^a}}} + \frac{{{x^c}}}{{{x^a}}}}} + \frac{1}{{1 + \frac{{{x^a}}}{{{x^b}}} + \frac{{{x^c}}}{{{x^b}}}}} + \frac{1}{{1 + \frac{{{x^b}}}{{{x^c}}} + \frac{{{x^a}}}{{{x^c}}}}} \cr} $$
  $$ = \frac{{{x^a}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$    $$\frac{{{x^b}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$   $$\frac{{{x^c}}}{{\left( {{x^a} + {x^{b}} + {x^c}} \right)}}$$
$$\eqalign{ & = \frac{{\left( {{x^a} + {x^b} + {x^c}} \right)}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} \cr & = 1 \cr} $$

$$\eqalign{ & {\text{We know that}} \cr & {\text{1}}{{\text{1}}^2} = 121 \cr & {\text{Putting}} \cr & m = 11\& n = 2 \cr & {\text{we get}} \cr & {\left( {m - 1} \right)^{n + 1}} \cr & = {\left( {11 - 1} \right)^{\left( {2 + 1} \right)}} \cr & = {10^3} \cr & = 1000 \cr} $$

$$1 + \left( {3 + 1} \right)$$  $$\left( {{3^2} + 1} \right)$$ $$\left( {{3^4} + 1} \right)$$ $$\left( {{3^8} + 1} \right)$$ $$\left( {{3^{16}} + 1} \right)$$ $$\left( {{3^{32}} + 1} \right)$$
$$ = 1 + \frac{1}{2}\left[ {\left( {3 - 1} \right)\left( {3 + 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$
$$ = 1 + \frac{1}{2}\left[ {\left( {{3^2} - 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$
$$ = 1 + \frac{1}{2}\left[ {\left( {{3^4} - 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right]$$
$$\eqalign{ & = 1 + \frac{1}{2}\left[ {\left( {{3^8} - 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{16}} - 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{32}} - 1} \right)\left( {{3^{32}} + 1} \right)} \right] \cr & = 1 + \frac{1}{2}\left[ {\left( {{3^{64}} + 1} \right)} \right] \cr & = \frac{{2 + {3^{64}} - 1}}{2} \cr & = \frac{{{3^{64}} + 1}}{2} \cr} $$

$$\eqalign{ & \frac{{\left( {x - \sqrt {24} } \right)\left( {\sqrt {75} + \sqrt {50} } \right)}}{{\sqrt {75} - \sqrt {50} }}{\text{ = 1 }} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{\sqrt {75} - \sqrt {50} }}{{\sqrt {75} + \sqrt {50} }}{\text{ }} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{{{\left( {\sqrt {75} - \sqrt {50} } \right)}^2}}}{{75 - 50}} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{75 + 50 - 2\sqrt {75} \sqrt {50} }}{{25}} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{125 - 2 \times 5\sqrt 3 \times 5\sqrt 2 }}{{25}} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{125 - 50\sqrt 6 }}{{25}} \cr & \Rightarrow \left( {x - \sqrt {24} } \right) = \frac{{25\left( {5 - 2\sqrt 6 } \right)}}{{25}} \cr & \Rightarrow x - 2\sqrt 6 = 5 - 2\sqrt 6 \cr & \Rightarrow x = 5{\text{ }} \cr} $$

$$\frac{{1 + \sqrt 2 }}{{\sqrt 5 + \sqrt 3 }} + \frac{{1 - \sqrt 2 }}{{\sqrt 5 - \sqrt 3 }}$$
$$ \Rightarrow \frac{{\left( {1 + \sqrt 2 } \right)\left( {\sqrt 5 - \sqrt 3 } \right) + \left( {1 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 3 } \right)}}{{\left( {\sqrt 5 + \sqrt 3 } \right)\left( {\sqrt 5 - \sqrt 3 } \right)}}$$
$$ \Rightarrow \frac{{\sqrt 5 - \sqrt 3 + \sqrt {10} - \sqrt 6 + \sqrt 5 + \sqrt 3 - \sqrt {10} - \sqrt 6 }}{{5 - 3}}$$
$$\eqalign{ & \Rightarrow \frac{{2\sqrt 5 - 2\sqrt 6 }}{2} \cr & \Rightarrow \frac{{2\left( {\sqrt 5 - \sqrt 6 } \right)}}{2} \cr & \Rightarrow \sqrt 5 - \sqrt 6 \cr} $$

Shortcut method:
When the question is in form
$$\eqalign{ & \Rightarrow \sqrt {n\sqrt {n\sqrt n } } ........\infty \cr & \Rightarrow {\text{So }}n{\text{ is answer }} \cr & \Rightarrow {\text{3}} \cr} $$

$$\eqalign{ & {\text{Let }}{25^{2.7}} \times {5^{4.2}} \div {5^{5.4}} = {25^x} \cr & {\text{Then, }}{25^{2.7}} \times {5^{(4.2 - 5.4)}} = {25^x} \cr & \Leftrightarrow {25^{2.7}} \times {5^{( - 1.2)}} = {25^x} \cr & \Leftrightarrow {25^{2.7}} \times \frac{1}{{{5^{1.2}}}} = {25^x} \cr & \Leftrightarrow \frac{{{{25}^{2.7}}}}{{{{\left( {{5^2}} \right)}^{0.6}}}} = {25^x} \cr & \Leftrightarrow \frac{{{{\left( {25} \right)}^{2.7}}}}{{{{\left( {25} \right)}^{0.6}}}} = {25^x} \cr & \Leftrightarrow {25^x} = {25^{\left( {2.7 - 0.6} \right)}} = {25^{2.1}} \cr & \Leftrightarrow x = 2.1 \cr} $$

$$\eqalign{ & {\text{Let }}\,{6^{1.2}} \times {36^x} \times {30^{2.4}} \times {25^{1.3}} = {30^5} \cr & {\text{Then,}}\,{6^{1.2}} \times {({6^2})^x} \times {(6 \times 5)^{2.4}} \times {({5^2})^{1.3}} = {30^5} \cr & \Leftrightarrow {6^{1.2}} \times {6^{2x}} \times {6^{2.4}} \times {5^{2.4}} \times {5^{2.6}} = {(6 \times 5)^5} \cr & \Leftrightarrow {6^{\left( {1.2 + 2x + 2.4} \right)}} \times {5^{\left( {2.4 + 2.6} \right)}} = {6^5} \times {5^5} \cr & \Leftrightarrow {6^{\left( {3.6 + 2x} \right)}} \times {5^5} = {6^5} \times {5^5} \cr & \Leftrightarrow 3.6 + 2x = 5 \cr & \Leftrightarrow 2x = 1.4 \cr & \Leftrightarrow x = 0.7 \cr} $$

$$\eqalign{ & {\left( {\sqrt 8 } \right)^{\frac{1}{3}}} \cr & = {\left( {{8^{\frac{1}{2}}}} \right)^{\frac{1}{3}}} \cr & = {8^{\left( {\frac{1}{2} \times \frac{1}{3}} \right)}} \cr & = {8^{\frac{1}{6}}} \cr & = {\left( {{2^3}} \right)^{\frac{1}{6}}} \cr & = {2^{\left( {3 \times \frac{1}{6}} \right)}} \cr & = {2^{\frac{1}{2}}} \cr & = \sqrt 2 \cr} $$

$$\left[ {\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} - \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}} \right]$$
$$ = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times $$   $$\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} - $$   $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times $$  $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$$
$$\eqalign{ & = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{3 - 2}} - \frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{3 - 2}} \cr & = \left( {3 + 2 + 2\sqrt 6 } \right) - \left( {3 + 2 - 2\sqrt 6 } \right) \cr & = 4\sqrt 6 {\text{ }} \cr} $$

$$\eqalign{ & {{\text{2}}^x} = \root 3 \of {32} \cr & \Rightarrow {{\text{2}}^x} = {\left( {32} \right)^{\frac{1}{3}}} = {\left( {{2^5}} \right)^{\frac{1}{3}}} \cr & \Rightarrow x = \frac{5}{3} \cr} $$

$$\sqrt {12 + \sqrt {12 + \sqrt {12 + ..... } } } $$
(3, 4) are the factor of 4
If there is '+' in $$\sqrt {\,\,\,} $$  Answer is Highest value
If there is '-' in $$\sqrt {\,\,\,} $$  Answer is lowest value.
So the answer is 4

$$\eqalign{ & {\text{1}}{{\text{0}}^{ - 8x}} \cr & = {\left[ {{{10}^x}} \right]^{ - 8}} \cr & = {\left[ {\frac{1}{2}} \right]^{ - 8}} \cr & = {2^8} \cr & = 256 \cr} $$

$$\left( {\frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + \frac{1}{{10.13}} + \frac{1}{{13.16}}} \right)$$
Formula :
$$\frac{1}{{{\text{Difference denominator value}}}} \times $$     $$\left[ {\frac{1}{{{\text{First value}}}} - \frac{1}{{{\text{Last value}}}}} \right]$$
$$ = \frac{1}{3} \times $$ $$\left[ {1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{10}} + \frac{1}{{10}} - \frac{1}{{13}} + \frac{1}{{13}} - \frac{1}{{16}}} \right]$$
$$\eqalign{ & = \frac{1}{3} \times \left[ {1 - \frac{1}{{16}}} \right] \cr & = \frac{1}{3} \times \frac{{15}}{{16}} \cr & = \frac{5}{{16}} \cr} $$