$$\eqalign{ & \frac{a}{3} = \frac{b}{2} \Rightarrow \frac{a}{b} = \frac{3}{2} \cr & \therefore \frac{{2a + 3b}}{{3a - 2b}} \cr & = \frac{{2 \times 3 + 3 \times 2}}{{3 \times 3 - 2 \times 2}} \cr & = \frac{{6 + 6}}{{9 - 4}} \cr & = \frac{{12}}{5} \cr} $$

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

$$\eqalign{ & {4^{4x + 1}} = \frac{1}{{64}} \cr & \Rightarrow {4^{4x + 1}} = \frac{1}{{{{\left( 4 \right)}^3}}} \cr & \Rightarrow {4^{4x + 1}} = {\left( 4 \right)^{ - 3}} \cr & \Rightarrow 4x + 1 = - 3 \cr & \Rightarrow 4x = - 4 \cr & \Rightarrow x = - 1 \cr} $$

$$\eqalign{ & {\left( {\sqrt 5 } \right)^7} \div {\left( {\sqrt 5 } \right)^5}{\text{ = }}{{\text{5}}^{\text{P}}} \cr & \Rightarrow \frac{{{{\left( {\sqrt 5 } \right)}^7}}}{{{{\left( {\sqrt 5 } \right)}^5}}} = {{\text{5}}^{\text{P}}} \cr & \Rightarrow {\left( {\sqrt 5 } \right)^2} = {{\text{5}}^{\text{P}}} \cr & \Rightarrow {{\text{5}}^{\text{1}}} = {\text{ }}{{\text{5}}^{\text{P}}} \cr & \Rightarrow \boxed{{\text{P}} = 1} \cr} $$

$$\eqalign{ & {\text{ }}n + \frac{2}{3}n + \frac{1}{2}n + \frac{1}{7}n = 97 \cr & \Rightarrow \frac{{42n + 28n + 21n + 6n}}{{42}} = 97 \cr & \Rightarrow \frac{{97n}}{{42}} = 97 \cr & \Rightarrow n = 42 \cr} $$

$$\eqalign{ & \Rightarrow x = 3 + \sqrt 8 \cr & \Rightarrow {x^2} = 9 + 8 + 2 \times 3\sqrt 8 \cr & \Rightarrow {x^2} = 17 + 6\sqrt 8 \cr & \Rightarrow \frac{1}{{{x^2}}} = 17 - 6\sqrt 8 \cr & \therefore {x^2} + \frac{1}{{{x^2}}} \cr & = 17 + 6\sqrt 8 + 17 - 6\sqrt 8 \cr & = 34 \cr} $$

$$\eqalign{ & x = 5 + 2\sqrt 6 \cr & \Leftrightarrow x = 3 + 2 + 2\sqrt 3 \times \sqrt 2 \cr & \Leftrightarrow x = {\left( {\sqrt 3 } \right)^2} + {\left( {\sqrt 2 } \right)^2} + 2\sqrt 3 \times \sqrt 2 \cr & \Leftrightarrow x = {\left( {\sqrt 3 + \sqrt 2 } \right)^2} \cr & \Leftrightarrow \sqrt x = \sqrt 3 + \sqrt 2 \cr & {\text{Similarly,}} \cr & \Leftrightarrow \frac{1}{{\sqrt x }} = \sqrt 3 - \sqrt 2 \cr & \therefore \left( {\sqrt x + \frac{1}{{\sqrt x }}} \right) \cr & = \sqrt 3 + \sqrt 2 + \sqrt 3 - \sqrt 2 \cr & = 2\sqrt 3 \cr} $$

$$\eqalign{ & 1.5a = 0.04b \cr & \frac{a}{b} = \frac{{0.04}}{{1.5}} = \frac{4}{{100}} \times \frac{{10}}{{15}} = \frac{2}{{75}} \cr & {\text{Let }}a = 2x,{\text{ }}b = 75x \cr & \therefore \frac{{b - a}}{{b + a}} = \frac{{75x - 2x}}{{75x + 2x}} = \frac{{73}}{{77}} \cr & \cr & {\bf{Alternate:}} \cr & \frac{a}{b} = \frac{{0.04}}{{1.5}} \cr & \therefore \frac{{b - a}}{{b + a}} = \frac{{1.5 - 0.04}}{{1.5 + 0.04}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1.46}}{{1.54}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{73}}{{77}} \cr} $$

$$\eqalign{ & \frac{{{x^2} - x + 1}}{{{x^2} + x + 1}} = \frac{3}{2}\left( {{\text{Given}}} \right) \cr & \Rightarrow \frac{{x\left\{ {\left( {x + \frac{1}{x}} \right) - 1} \right\}}}{{x\left\{ {\left( {x + \frac{1}{x}} \right) + 1} \right\}}} = \frac{3}{2} \cr & \Rightarrow \frac{{\left( {x + \frac{1}{x}} \right) - 1}}{{\left( {x + \frac{1}{x}} \right) + 1}} = \frac{3}{2} \cr & \,\,\,\,\,\,\,\,\,\,{\text{Let }}\left( {x + \frac{1}{x} = y} \right) \cr & \Rightarrow \frac{{y - 1}}{{y + 1}} = \frac{3}{2} \cr & \Rightarrow 2\left( {y - 1} \right) = 3\left( {y + 1} \right) \cr & \Rightarrow 2y - 2 = 3y + 3 \cr & \Rightarrow y = - 2 - 3 \cr & \Rightarrow y = - 5 \cr & \therefore x + \frac{1}{x} = - 5 \cr & {\text{ }} \cr} $$

$$\eqalign{ & \frac{{\sqrt 7 - 2}}{{\sqrt 7 + 2}} = a\sqrt 7 + b \cr & {\text{L}}{\text{.H}}{\text{.S }}\frac{{\sqrt 7 - 2}}{{\sqrt 7 + 2}} \times \frac{{\sqrt 7 - 2}}{{\sqrt 7 - 2}} \cr & {\text{ }}\left( {{\text{Rationalisation}}} \right) \cr & = \frac{{{{\left( {\sqrt 7 - 2} \right)}^2}}}{{{{\left( {\sqrt 7 } \right)}^2} - \left( 4 \right)}} \cr & = \frac{{7 + 4 - 4\sqrt 7 }}{{7 - 4}} \cr & = \frac{{11 - 4\sqrt 7 }}{3} \cr & = \frac{{11}}{3} - \frac{4}{3}\sqrt 7 \cr & = - \frac{4}{3}\sqrt 7 + \frac{{11}}{3} \cr & = a\sqrt 7 + b{\text{ }}\left( {{\text{ R}}{\text{.H}}{\text{.S}}{\text{.}}} \right) \cr} $$
(Compare the coefficients of $$\sqrt 7 $$ and constant term)
$$\eqalign{ & {\text{a}} = - \frac{4}{3} \cr & b = \frac{{11}}{3} \cr} $$

$$\eqalign{ & a + \frac{1}{b} = 1{\text{, }}b + \frac{1}{c} = 1{\text{, }}c + \frac{1}{a} = ? \cr & {\text{Put values, }} \cr & a = \frac{1}{2},b = 2,{\text{ }}c = - 1 \cr & c + \frac{1}{a} \cr & = - 1 + \frac{1}{{\left( {\frac{1}{2}} \right)}} \cr & = - 1 + 2 \cr & = 1 \cr & \cr & {\bf{Alternate:}} \cr & \Rightarrow a + \frac{1}{b} = 1 \cr & \Rightarrow a = 1 - \frac{1}{b} = \boxed{\frac{{b - 1}}{b}} \cr & \frac{1}{a} = \frac{b}{{b - 1}} \cr & \Rightarrow b + \frac{1}{c} = 1 \cr & \Rightarrow \frac{1}{c} = 1 - b,\boxed{c = \frac{1}{{1 - b}}} \cr & \therefore c + \frac{1}{a} \cr & = \frac{1}{{1 - b}} + \frac{b}{{b - 1}} \cr & = \frac{1}{{1 - b}} - \frac{b}{{1 - b}} \cr & = \frac{{1 - b}}{{1 - b}} \cr & = 1 \cr} $$

$$\eqalign{ & x - \frac{1}{x} = 4 \cr & \left( {{\text{on squaring }}} \right) \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 16 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 18 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 - 2 = 18 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 20 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 20 \cr & \Rightarrow x + \frac{1}{x} = \sqrt {20} \cr & \Rightarrow x + \frac{1}{x} = \sqrt {4 \times 5} \cr & \Rightarrow x + \frac{1}{x} = 2\sqrt 5 \cr} $$

$$\eqalign{ & 4{b^2} + \frac{1}{{{b^2}}} = 2 \cr & \Rightarrow {\left( {2b} \right)^2} + {\left( {\frac{1}{b}} \right)^2} + 4 - 4 = 2 \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^2} - 4 = 2 \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^2} = 6 \cr & \Rightarrow 2b + \frac{1}{b} = \sqrt 6 \cr & {\text{Take cube both sides}} \cr & \Rightarrow {\left( {2b + \frac{1}{b}} \right)^3} = {\left( {\sqrt 6 } \right)^3} \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} + 3 \times 2b \times \frac{1}{b}\left( {2b + \frac{1}{b}} \right) = 6\sqrt 6 \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} + 6\sqrt 6 = 6\sqrt 6 \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} = 6\sqrt 6 - 6\sqrt 6 \cr & \Rightarrow 8{b^3} + \frac{1}{{{b^3}}} = 0 \cr} $$

$$\eqalign{ & {\text{ }}{x^{\frac{1}{3}}} = {y^{\frac{1}{4}}}, \cr & \Rightarrow {\text{LCM of 3, 4}} = 12{\text{ }} \cr & \therefore {\text{ }}{\left( {{x^{\frac{1}{3}}}} \right)^{12}} = {\left( {{y^{\frac{1}{4}}}} \right)^{12}} \cr & \Rightarrow {x^4} = {y^3} \cr & {\text{take power '5' on both sides}} \cr & \Rightarrow {\left( {{x^4}} \right)^5} = {\left( {{y^3}} \right)^5} \cr & \Rightarrow {x^{20}} = {y^{15}} \cr} $$

$$\eqalign{ & x = \frac{{\sqrt 3 }}{2} \cr & {\text{or }}1 + x = 1 + \frac{{\sqrt 3 }}{2} \cr & \Rightarrow 1 + x = \frac{{2 + \sqrt 3 }}{2} \cr & \Rightarrow 1 + x = \frac{{2\left( {2 + \sqrt 3 } \right)}}{{2 \times 2}} \cr & \left( {{\text{Divided and multiply by 2}}} \right) \cr & \Rightarrow 1 + x = \frac{{4 + 2\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{1 + 3 + 2\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{4 + 2\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{{{\left( 1 \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} + 2.1.\sqrt 3 }}{4} \cr & \Rightarrow 1 + x = \frac{{{{\left( {1 + \sqrt 3 } \right)}^2}}}{4} \cr & \therefore \sqrt {1 + x} = \frac{{1 + \sqrt 3 }}{2} \cr & \cr & {\bf{Similarly:}} \cr & \sqrt {1 - x} \cr & = \frac{{\sqrt 3 - 1}}{2} \cr & \therefore \frac{{\sqrt {1 + x} }}{{1 + \sqrt {1 + x} }}{\text{ + }}\frac{{\sqrt {1 - x} }}{{1 - \sqrt {1 - x} }} \cr & = \frac{{\frac{{1 + \sqrt 3 }}{2}}}{{1 + \frac{{1 + \sqrt 3 }}{2}}} + \frac{{\frac{{\sqrt 3 - 1}}{2}}}{{1 - \frac{{\sqrt 3 - 1}}{2}}} \cr & = \frac{{1 + \sqrt 3 }}{{3 + \sqrt 3 }} + \frac{{\sqrt 3 - 1}}{{3 - \sqrt 3 }} \cr & = \frac{{1 + \sqrt 3 }}{{\sqrt 3 \left( {\sqrt 3 + 1} \right)}} + \frac{{1 - \sqrt 3 }}{{\sqrt 3 \left( {\sqrt 3 - 1} \right)}} \cr & = \frac{1}{{\sqrt 3 }} + \frac{1}{{\sqrt 3 }} \cr & = \frac{2}{{\sqrt 3 }} \cr} $$

$$\eqalign{ & x + \frac{9}{x} = 6 \cr & {\text{Take value of x}} \cr & {\text{Let }}x = 3 \cr & 3 + \frac{9}{3} = 6{\text{ }}\left( {{\text{Proved}}} \right) \cr & {\text{So, }}x = 3 \cr & \therefore {x^2} + \frac{9}{{{x^2}}} \cr & = 9 + \frac{9}{9} \cr & = 10 \cr & \cr & {\bf{Alternate:}} \cr & x + \frac{9}{x} = 6 \cr & {\text{On squaring,}} \cr & \Rightarrow {\left( {x + \frac{9}{x}} \right)^2} = 36 \cr & \Rightarrow {x^2} + \frac{{81}}{{{x^2}}} + 2 \times x \times \frac{9}{x} = 36 \cr & \Rightarrow {x^2} + \frac{{81}}{{{x^2}}} - 18 = 0 \cr & \Rightarrow {\left( {x - \frac{9}{x}} \right)^2} = 0 \cr & \Rightarrow x = \frac{9}{x} \cr & \Rightarrow {x^2} = 9 \cr & {\text{Hence, }}\left( {{x^2} + \frac{9}{{{x^2}}}} \right) \cr & = \left( {9 + \frac{9}{9}} \right) \cr & = 10 \cr} $$

$$\eqalign{ & \frac{{{{\left( {x + y + z} \right)}^2}}}{{{x^2} + {y^2} + {z^2}}} \cr & {\text{Assume }}x = y = z = 1 \cr & \Leftrightarrow \frac{{{{\left( {1 + 1 + 1} \right)}^2}}}{{1 + 1 + 1}} \cr & \Leftrightarrow \frac{9}{3} \cr & \Leftrightarrow 3 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 3{\text{ }}\left( {{\text{Given}}} \right) \cr & \frac{{{x^3} + \frac{1}{x}}}{{{x^2} - x + 1}}{\text{ }}\left( {{\text{Divide by }}x} \right) \cr & \Rightarrow \frac{{\frac{{{x^3}}}{x} + \frac{1}{{{x^2}}}}}{{\frac{{{x^2}}}{x} - \frac{x}{x} + \frac{1}{x}}} \cr & \Rightarrow \frac{{{x^2} + \frac{1}{{{x^2}}}}}{{x - 1 + \frac{1}{x}}} \cr & \Rightarrow \frac{{{x^2} + \frac{1}{{{x^2}}}}}{{x + \frac{1}{x} - 1}} \cr & \therefore x + \frac{1}{x} = 3 \cr & \therefore {x^2} + \frac{1}{{{x^2}}} = 9 - 2 = 7 \cr & \therefore \frac{{{x^2} + \frac{1}{{{x^2}}}}}{{x + \frac{1}{x} - 1}} = \frac{7}{{3 - 1}} = \frac{7}{2} \cr} $$

$$\eqalign{ & {a^2} + {b^2} + {c^2} - ab - bc - ca \cr & = \frac{1}{2}\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \cr & = \frac{1}{2}\left[ {{{\left( {7 - 5} \right)}^2} + {{\left( {5 - 3} \right)}^2} + {{\left( {3 - 7} \right)}^2}} \right] \cr & = \frac{1}{2}\left( {4 + 4 + 16} \right) \cr & = \frac{{24}}{2} \cr & = 12 \cr} $$

$$\eqalign{ & {{\text{7}}^x}{\text{ = }}\frac{1}{{343}} \cr & \Rightarrow {{\text{7}}^x}{\text{ = }}\frac{1}{{{7^3}}} \cr & \Rightarrow {{\text{7}}^x}{\text{ = }}{{\text{7}}^{ - 3}} \cr & \Rightarrow x = - 3 \cr} $$
(If bases are equal then their power are also equal)

$$\eqalign{ & x + y = 7 \cr & \left( {{\text{cubing both sides}}} \right){\text{ }} \cr & \Rightarrow {\left( {x + y} \right)^3} = {\left( 7 \right)^3} \cr & \Rightarrow {x^3} + {y^3} + 3\left( {x + y} \right)xy = 343 \cr & \Rightarrow {x^3} + {y^3} + 21xy = 343 \cr} $$

$$\eqalign{ & {\text{Given, }}a \otimes b = 2a{\text{ }} \cr & {\text{Where }}a > b \cr & a \otimes b = a + b \cr & {\text{Where }}a < b \cr & a \otimes b = {a^2} \cr & {\text{Where }}a = b \cr & \frac{{\left( {5 \otimes 7} \right) + \left( {4 \otimes 4} \right)}}{{3\left( {5 \otimes 5} \right) - \left( {15 \otimes 11} \right) - 3}}{\text{ }} \cr & {\text{ = }}\frac{{\left( {5 + 7} \right) + {{\left( 4 \right)}^2}}}{{3{{\left( 5 \right)}^2} - \left( {2 \times 15} \right) - 3}} \cr & {\text{ = }}\frac{{12 + 16}}{{75 - 30 - 3}} \cr & {\text{ = }}\frac{{28}}{{42}} \cr & {\text{ = }}\frac{2}{3} \cr} $$

$$\eqalign{ & {\left( {\frac{3}{5}} \right)^3} \times {\left( {\frac{3}{5}} \right)^{ - 6}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & \Rightarrow {\left( {\frac{3}{5}} \right)^{3 - 6}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & \Rightarrow - 3 = 2x - 1 \cr & \Rightarrow 2x = - 2 \cr & \Rightarrow x = - 1 \cr} $$

$$\eqalign{ & \frac{{\sqrt {3 + x} + \sqrt {3 - x} }}{{\sqrt {3 + x} - \sqrt {3 - x} }} = \frac{2}{1}{\text{ }} \cr & \Rightarrow \frac{{\sqrt {3 + x} }}{{\sqrt {3 - x} }} = \frac{{2 + 1}}{{2 - 1}} = \frac{3}{1} \cr & \left[ {\frac{{\text{A}}}{{\text{B}}} = \frac{{\text{C}}}{{\text{D}}}} \right] \cr & \left[ {\frac{{{\text{A}} + {\text{B}}}}{{{\text{A}} - {\text{B}}}} = \frac{{{\text{C}} + {\text{D}}}}{{{\text{C}} - {\text{D}}}}} \right] \cr & \Rightarrow \frac{{\sqrt {3 + x} }}{{\sqrt {3 - x} }} = 3{\text{ }} \cr & {\text{Squaring both sides}} \cr & \Rightarrow \frac{{3 + x}}{{3 - x}} = 9 \cr & \Rightarrow 3 + x = 27 - 9x \cr & \Rightarrow 10x = 24 \cr & \Rightarrow x = \frac{{24}}{{10}} \cr & \Rightarrow x = \frac{{12}}{5} \cr} $$

$$\eqalign{ & \Leftrightarrow {x^2} - 4x - 1 = 0 \cr & \Leftrightarrow {x^2} - 1 = 4x \cr & \left( {{\text{divide }}x{\text{ both sides}}} \right) \cr & \Leftrightarrow x - \frac{1}{x} = 4 \cr & \Leftrightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 16 \cr & \Leftrightarrow {x^2} + \frac{1}{{{x^2}}} = 18 \cr} $$

$$\left( {x + \frac{1}{x}} \right)$$ $$\left( {x - \frac{1}{x}} \right)$$ $$\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)$$  $$\left( {{x^2} + \frac{1}{{{x^2}}} + 1} \right)$$
$${\text{ = }}\left( {x + \frac{1}{x}} \right)$$ $$\left( {{x^2} + \frac{1}{{{x^2}}} - 1} \right)$$  $$\left( {x - \frac{1}{x}} \right)$$ $$\left( {{x^2} + \frac{1}{{{x^2}}} + 1} \right)$$
$$\eqalign{ & \because \left( {{\text{A}} + {\text{B}}} \right)\left( {{{\text{A}}^2} - {\text{AB}} + {{\text{B}}^2}} \right) = {{\text{A}}^3} + {{\text{B}}^3} \cr & \because \left( {{\text{A}} - {\text{B}}} \right)\left( {{{\text{A}}^2} + {\text{AB}} + {{\text{B}}^2}} \right) = {{\text{A}}^3} - {{\text{B}}^3} \cr & = \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^3} - \frac{1}{{{x^3}}}} \right) \cr & = \boxed{{x^6} - \frac{1}{{{x^6}}}} \cr} $$

$$\eqalign{ & a = 4.36 \cr & b = 2.39 \cr & c = 1.97 \cr & {\text{ }}a - b - c \cr & = 4.36 - 2.39 - 1.97 \cr & = 0 \cr & {\text{ }}{{\text{a}}^3} - {b^3} - {c^3} - 3abc \cr & = \frac{1}{2}\left( {a - b - c} \right)\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \cr & = 0 \cr} $$

$$\eqalign{ & a = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \cr & b = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}}{\text{ }} \cr & \therefore a = \frac{1}{b} \cr & a + b = a + \frac{1}{a} \cr & \Rightarrow \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ + }}\frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \Rightarrow \frac{{5 + 1 + 2\sqrt 5 + 5 + 1 - 2\sqrt 5 }}{{{{\left( {\sqrt 5 } \right)}^2} - {{\left( 1 \right)}^2}}} \cr & \Rightarrow \frac{{6 + 2\sqrt 5 + 6 - 2\sqrt 5 }}{{5 - 1}} \cr & \Rightarrow \frac{{12}}{4} \cr & \Rightarrow 3 \cr & \therefore \frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}} \cr & \Rightarrow \frac{{{a^2} + \frac{1}{{{a^2}}} + ab}}{{{a^2} + \frac{1}{{{a^2}}} - ab}} \cr & \Rightarrow a + \frac{1}{a} = 3 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} \cr & \Rightarrow 9 - 2 \cr & \Rightarrow 7\left( {ab = 1} \right) \cr & \therefore \frac{{{a^2} + \frac{1}{{{a^2}}} + ab}}{{{a^2} + \frac{1}{{{a^2}}} - ab}} \cr & = \frac{{7 + 1}}{{7 - 1}} \cr & = \frac{8}{6} \cr & = \frac{4}{3} \cr} $$

$$\eqalign{ & {\text{If }}a + b + c = 0 \cr & {\text{Put }}a = 1,{\text{ }}b = 1{\text{ and }}c = - 2 \cr & \therefore \frac{{{a^2} + {b^2} + {c^2}}}{{ab + bc + ca}} \cr & = \frac{{1 + 1 + 4}}{{1 - 2 - 2}} \cr & = \frac{6}{{ - 3}} \cr & = - 2 \cr} $$

$$\eqalign{ & a + b + c = m \cr & \Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0 \cr & \Rightarrow \frac{{ab + bc + ca}}{{abc}} = 0 \cr & \Rightarrow ab + bc + ca = 0 \cr & {\left( {a + b + c} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) \cr & \Rightarrow {m^2} = {\text{ }}{a^2} + {b^2} + {c^2} \cr & \Rightarrow \frac{{{m^2}}}{3} = {\text{ }}\frac{{{a^2} + {b^2} + {c^2}}}{3} \cr} $$