Solution:
$$\eqalign{ a - \frac{1}{{a - 3}} = 5 \cr \Rightarrow a - 3 - \frac{1}{{a - 3}} = 5 - 3 \cr \Rightarrow \left( {a - 3} \right) - \frac{1}{{\left( {a - 3} \right)}} = 2 \cr {\text{Cubing both sides}} \cr \Rightarrow {\left[ {\left( {a - 3} \right) - \frac{1}{{\left( {a - 3} \right)}}} \right]^3} = {\left( 2 \right)^3} \cr} $$
  $$ \Rightarrow {\left( {a - 3} \right)^3} - \frac{1}{{{{\left( {a - 3} \right)}^3}}} - 3 \times {\left( {a - 3} \right)} \times $$       $$\frac{1}{{{{\left( {a - 3} \right)}}}}$$ $$\left[ {\left( {a - 3} \right) - \frac{1}{{\left( {a - 3} \right)}}} \right]$$    $$ = 8$$
$$\eqalign{ \Rightarrow {\left( {a - 3} \right)^3} - \frac{1}{{{{\left( {a - 3} \right)}^3}}} - 3\left( 2 \right) = 8 \cr \Rightarrow {\left( {a - 3} \right)^3} - \frac{1}{{{{\left( {a - 3} \right)}^3}}} = 8 + 6 \cr \Rightarrow {\left( {a - 3} \right)^3} - \frac{1}{{{{\left( {a - 3} \right)}^3}}} = 14 \cr} $$

Solution:
$$\eqalign{ x + \frac{1}{x} = 5{\text{ then, }}\frac{{5x}}{{{x^2} + 5x + 1}} \cr \Rightarrow \frac{5}{{x + 5 + \frac{1}{x}}} \cr \Rightarrow \frac{5}{{x + \frac{1}{x} + 5}} \cr \Rightarrow \frac{5}{{5 + 5}} \cr \Rightarrow \frac{1}{2} \cr} $$

Solution:
$$\eqalign{ {\left( {x + \frac{1}{x}} \right)^2} = 3 \cr \Rightarrow x + \frac{1}{x} = \sqrt 3 \cr \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 0 \cr \Rightarrow {x^6} + 1 = 0 \cr \Rightarrow {x^6} = - 1 \cr \Rightarrow {x^{72}} + {x^{66}} + {x^{54}} + {x^{24}} + {x^6} + 1 \cr \Rightarrow {\left( {{x^6}} \right)^{12}} + {\left( {{x^6}} \right)^{11}} + {\left( {{x^6}} \right)^9} + {\left( {{x^6}} \right)^4} + {x^6} + 1 \cr \Rightarrow {\left( { - 1} \right)^{12}} + {\left( { - 1} \right)^{11}} + {\left( { - 1} \right)^9} + {\left( { - 1} \right)^4} - 1 + 1 \cr \Rightarrow 1 - 1 - 1 + 1 - 1 + 1 \cr \Rightarrow 0 \cr} $$

Solution:
$$\eqalign{ {\text{Given, }}x + \frac{1}{x} = 1 \cr {\text{Find }}\frac{{{x^2} + 3x + 1}}{{{x^2} + 7x + 1}} \cr {\text{From equation (i)}} \cr \Rightarrow x + \frac{1}{x} = 1 \cr \Rightarrow {x^2} + 1 = x \cr \Rightarrow \frac{{\left( {{x^2} + 1} \right) + 3x}}{{\left( {{x^2} + 1} \right) + 7x}} \cr \Rightarrow \frac{{x + 3x}}{{x + 7x}} \cr \Rightarrow \frac{{4x}}{{8x}} \cr \Rightarrow \frac{1}{2} \cr} $$

Solution:
$$\eqalign{ {x^2} - 3x + 1 = 0 \cr {x^2} + 1 = 3x \cr x + \frac{1}{x} = 3 \cr \Rightarrow \frac{{\left( {{x^4} + \frac{1}{{{x^2}}}} \right)}}{{\left( {{x^2} + 5x + 1} \right)}} \cr = \frac{{x\left( {{x^3} + \frac{1}{{{x^3}}}} \right)}}{{\left( {{x^2} + 1 + 5x} \right)}} \cr = \frac{{x\left[ {{3^3} - 3 \times 3} \right]}}{{\left( {3x + 5x} \right)}} \cr = \frac{{x\left[ {18} \right]}}{{\left( {8x} \right)}} \cr = \frac{9}{4} \cr} $$

Solution:
$$\left[ {\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\, - \,\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,} \right]$$              $$ ÷ $$  $$\left[ {\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\, + \,\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,} \right]$$
$$ {\text{Let }}\left[ {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right] = a,$$    $${\text{ }}\left[ {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right] = b $$
$$\eqalign{ \Rightarrow \left( {{a^2} - {b^2}} \right) \div \left( {a + b} \right) = a - b = ? \cr \Rightarrow a = 1 + \frac{{10}}{{101}} \cr \Rightarrow a = \frac{{111}}{{101}} \cr \Rightarrow b = 1 - \frac{{10}}{{101}} \cr \Rightarrow b = \frac{{91}}{{101}} \cr \Rightarrow a - b = \frac{{111}}{{101}} - \frac{{91}}{{101}} \cr \Rightarrow a - b = \frac{{20}}{{101}} \cr} $$

Solution:
If (x - 1) and (x + 1) are the factors y equation then,
$$\eqalign{ x - 1 = 0 \cr x = 1 \cr \Rightarrow {\text{Put }}x = 1{\text{, we get }} \cr 1 + a - 3 + 2 + b = 0 \cr a + b = 0\,.....(i) \cr \Rightarrow x + 1 = 0 \cr \Rightarrow x = - 1 \cr {\text{Put }}x = - 1,{\text{we get }} \cr 1 - a - 3 - 2 + b = 0 \cr b - a = 4\,.....(ii) \cr {\text{After solving (i) (ii),}} \cr {\text{We get }} \cr a = - 2,{\text{ }}b = 2 \cr} $$