$$\eqalign{ & {\text{Put }}a = b = c = 1 \cr & {\text{Then we have }} \cr & \frac{{m - 3}}{2}{\text{ + }}\frac{{m - 3}}{2}{\text{ + }}\frac{{m - 3}}{2} = 9{\text{ }} \cr & m = 9 \cr} $$
Now, putting values of a, b, c in option,
Only option (C) gives value of m = 9
So, option (C) is correct.

$$\eqalign{ & a + \frac{1}{a} + 1 = 0 \cr & a + \frac{1}{a} = - 1 \cr & {\text{Squaring both sides}} \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} + 2 = 1 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} = - 1 \cr & \Rightarrow {a^2} + 1 = - \frac{1}{{{a^2}}}\,.....(i) \cr & \Rightarrow a + \frac{1}{a} = - 1{\text{ }}\left( {{\text{Given}}} \right) \cr & \therefore {a^2} + 1 = - a\,.....(ii) \cr & \Rightarrow - a = \frac{{ - 1}}{{{a^2}}} \cr & {\text{For equation (i) and (ii)}} \cr & {{\text{a}}^3} = 1 \cr & \therefore {{\text{a}}^3} - 1 = 0 \cr & \Rightarrow {a^4} - a = a\left( {{a^3} - 1} \right) \cr & \Rightarrow {a^4} - a = a \times 0 \cr & \Rightarrow {a^4} - a = 0 \cr} $$

$$\frac{{\frac{1}{3}.\frac{1}{3}.\frac{1}{3} + \frac{1}{4}.\frac{1}{4}.\frac{1}{4} - 3.\frac{1}{3}.\frac{1}{4}.\frac{1}{5} + \frac{1}{5}.\frac{1}{5}.\frac{1}{5}}}{{\frac{1}{3}.\frac{1}{3} + \frac{1}{4}.\frac{1}{4} + \frac{1}{5}.\frac{1}{5} - \left( {\frac{1}{3}.\frac{1}{4} + \frac{1}{4}.\frac{1}{5} + \frac{1}{5}.\frac{1}{3}} \right)}}$$

A³ + B³ + C³ - 3ABC = (A + B + C)(A² + B² + C² - AB - BC - CA)

$$\therefore \frac{{{{\left( {\frac{1}{3}} \right)}^3} + {{\left( {\frac{1}{4}} \right)}^3} - 3.\frac{1}{3}.\frac{1}{4}.\frac{1}{5} + {{\left( {\frac{1}{5}} \right)}^3}}}{{{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{4}} \right)}^2} + {{\left( {\frac{1}{5}} \right)}^2} - \frac{1}{3}.\frac{1}{4} - \frac{1}{4}.\frac{1}{5} - \frac{1}{5}.\frac{1}{3}}}$$
$$ = \frac{{\left( {\frac{1}{3} + \frac{1}{4} + \frac{1}{5}} \right)\left[ {{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{4}} \right)}^2} + {{\left( {\frac{1}{5}} \right)}^2} - \frac{1}{3}.\frac{1}{4} - \frac{1}{4}.\frac{1}{5} - \frac{1}{5}.\frac{1}{3}} \right]}}{{\left[ {{{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{4}} \right)}^2} + {{\left( {\frac{1}{5}} \right)}^2} - \frac{1}{3}.\frac{1}{4} - \frac{1}{4}.\frac{1}{5} - \frac{1}{5}.\frac{1}{3}} \right]}}$$
$$\eqalign{ & = \frac{{20 + 15 + 12}}{{60}} \cr & = \frac{{47}}{{60}} \cr} $$

$$\eqalign{ & {\text{0}}{\text{.13}} \times {{\text{p}}^2} = 13 \cr & \Rightarrow {{\text{p}}^2} = \frac{{13}}{{0.13}} \cr & \Rightarrow {{\text{p}}^2} = \frac{{13}}{{13}} \times 100 \cr & \Rightarrow {{\text{p}}^2} = 100 \cr & \Rightarrow {\text{p}} = 10 \cr} $$

$$\eqalign{ & {\left( {125} \right)^x} = 3125 \cr & \Rightarrow {\left( {{5^3}} \right)^x} = {\left( 5 \right)^5} \cr & \Rightarrow {\left( 5 \right)^{3x}} = {\left( 5 \right)^5} \cr & \Rightarrow 3x = 5 \cr & \Rightarrow x = \frac{5}{3} \cr} $$

$$\eqalign{ & \Rightarrow {a^{2x + 2}} = 1 \cr & \Rightarrow {a^{2x + 2}} = {a^0} \cr & \Rightarrow 2x + 2 = 0 \cr & \Rightarrow x = - \frac{2}{2} \cr & \Rightarrow x = - 1 \cr} $$

$$\eqalign{ & {\left( {a + \frac{1}{a}} \right)^2} = 3 \cr & \Rightarrow a + \frac{1}{a} = \sqrt 3 \cr & {\text{Cube both sides}} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3 \times a \times \frac{1}{a}\left( {a + \frac{1}{a}} \right) = {\left( {\sqrt 3 } \right)^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\left( {\sqrt 3 } \right) = 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr & \Rightarrow {a^6} + 1 = 0 \cr & \therefore {\text{ }}{a^{30}} + {a^{24}} + {a^{18}} + {a^{12}} + {a^6} + 1 = ? \cr & \Rightarrow {\text{ }}{a^{24}}\left( {{a^6} + 1} \right) + {a^{12}}\left( {{a^6} + 1} \right) + {a^6} + 1 \cr & \Rightarrow {a^{24}}\left( 0 \right) + {a^{12}}\left( 0 \right) + 0 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & \left( {\sqrt a + \sqrt b } \right) = 15{\text{ }} \cr & {\text{Square both sides}} \cr & \Rightarrow a + b + 2\sqrt {ab} = 225 \cr & \Rightarrow a + b = 225 - 2\sqrt {ab} \,.....(i) \cr & {\text{ }}\left( {\sqrt a - \sqrt b } \right) = 3 \cr & {\text{Square both sides}} \cr & \Rightarrow a + b - 2\sqrt {ab} = 9 \cr & \Rightarrow a + b = 9 + 2\sqrt {ab} \,.....(ii) \cr & {\text{From (i) and (ii)}} \cr & \Rightarrow 225 - 2\sqrt {ab} = 9 + 2\sqrt {ab} \cr & \Rightarrow 216 = 4\sqrt {ab} \cr & \Rightarrow 54 = \sqrt {ab} \cr & {\text{Divided by 4 on both sides}} \cr & \Rightarrow \frac{{\sqrt {ab} }}{4} = \frac{{54}}{4} \cr & \Rightarrow \frac{{\sqrt {ab} }}{4} = \frac{{27}}{2} \cr} $$

$$\eqalign{ & {\text{Given , }}a + \frac{1}{a} = 3 \cr & {\text{Cube both sides}} \cr & {a^3} + \frac{1}{{{a^3}}} + 3 \times a \times \frac{1}{a}\left( {a + \frac{1}{a}} \right) = {\left( 3 \right)^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 27 - 9 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 18 \cr} $$

$$\eqalign{ & {\text{ }}{x^2}{\text{ + }}\alpha x + \beta = 0 \cr & {\text{Sum of root}} \cr & \alpha + \beta = \frac{{ - \alpha }}{1}\,......(i) \cr & \alpha \beta = \beta \,......(ii) \cr & {\text{From (i) and (ii)}} \cr & {\text{Then, }}\alpha = 1 \cr & {\text{Then, }}\beta = - 2 \cr & {\text{Then value of }} \cr & \Leftrightarrow {\alpha ^3} + {\beta ^3} \cr & = 1 + {\left( { - 2} \right)^3} \cr & = - 7 \cr} $$

$$\eqalign{ & {a^2} = b + c,{\text{ }}{b^2} = a + c,{\text{ }}{c^2} = b + a \cr & {\text{Taking }}a = 2,{\text{ }}b = 2{\text{ and }}c = 2 \cr & {\text{So,}}{\left( 2 \right)^2} = 2 + 2 \cr & \boxed{4 = 4} \cr & {\text{Now,}}\frac{1}{{a + 1}} + {\text{ }}\frac{1}{{b + 1}} + {\text{ }}\frac{1}{{c + 1}} \cr & {\text{Put }}a = 2,{\text{ }}b = 2{\text{ and }}c = 2 \cr & = \frac{1}{{2 + 1}} + {\text{ }}\frac{1}{{2 + 1}} + {\text{ }}\frac{1}{{2 + 1}} \cr & = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} \cr & = 1 \cr} $$

$$\eqalign{ & x = 7 - 4\sqrt 3 \cr & \Rightarrow x = 4 + 3 - 4\sqrt 3 \cr & \Rightarrow x = {\left( 2 \right)^2} + {\left( {\sqrt 3 } \right)^2} - 2 \times 2\sqrt 3 \cr & \Rightarrow {\left( {2 - \sqrt 3 } \right)^2} \cr & \therefore \left[ {{a^2} + {b^2} - 2ab = {{\left( {a - b} \right)}^2}} \right] \cr & \Rightarrow x = {\left( {2 - \sqrt 3 } \right)^2} \cr & \Rightarrow \sqrt x = 2 - \sqrt 3 \cr & \Rightarrow \frac{1}{{\sqrt x }} = \frac{1}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} \cr & \Rightarrow 2 + \sqrt 3 \cr & \therefore \sqrt x {\text{ + }}\frac{1}{{\sqrt x }}{\text{ }} \cr & = 2 - \sqrt 3 + 2 + \sqrt 3 \cr & = 4 \cr} $$

$$\eqalign{ & x:y = 3:4 \cr & \frac{{\left( {7x + 3y} \right)}}{{\left( {7x - 3y} \right)}} \cr & = \frac{y}{y}\left( {\frac{{7\frac{x}{y} + 3}}{{7\frac{x}{y} - 3}}} \right) \cr & = \frac{{7 \times \frac{3}{4} + 3}}{{7 \times \frac{3}{4} - 3}} \cr & = \frac{{\frac{{21}}{4} + 3}}{{\frac{{21}}{4} - 3}} \cr & = \frac{{\frac{{21 + 12}}{4}}}{{\frac{{21 - 12}}{4}}} \cr & = \frac{{11}}{3} \,\,or\,\, 11 : 3 \cr} $$

$$\eqalign{ & x + \frac{1}{4}\sqrt x + {a^2} \cr & = {\left( {\sqrt x } \right)^2} + 2 \times \frac{1}{8} \times \sqrt x + {a^2} \cr & \left[ {\left( {{{\text{A}}^2} + {\text{2AB}} + {{\text{B}}^2}} \right) = {{\left( {{\text{A}} + {\text{B}}} \right)}^2}} \right] \cr & {\text{Here, A}} = \sqrt x {\text{ and }} \cr & {\text{B}} = a \cr & {\text{B}} = \frac{1}{8} \cr & \therefore a = \frac{1}{8} \cr} $$

$$\eqalign{ & x:y = 3:5{\text{ }} \cr & x - y = - 2 \cr & \frac{x}{y} = \frac{3}{5} \cr & x - y = 3 - 5 \cr & \Leftrightarrow - 2 = - 2 \cr & x = 3,{\text{ }}y = 5 \cr & x + y = 3 + 5 \cr & \Leftrightarrow x + y = 8 \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & x = 1 + \sqrt 2 + \sqrt 3 \,.....(i) \cr & y = 1 + \sqrt 2 - \sqrt 3 \,.....(ii) \cr & \Rightarrow \frac{{{x^2} + 4xy + {y^2}}}{{x + y}} \cr & \Rightarrow \frac{{{{\left( {x + y} \right)}^2} + 2xy}}{{x + y}} \cr & {\text{From equation (i)}} + {\text{(ii)}} \cr & \Rightarrow x + y = 2 + 2\sqrt 2 \cr & xy = {\left( {1 + \sqrt 2 } \right)^2} - {\left( {\sqrt 3 } \right)^2} \cr & \Rightarrow xy = 3 + 2\sqrt 2 - 3 \cr & \Rightarrow xy = 2\sqrt 2 \cr & {\text{So, }}\frac{{{{\left( {x + y} \right)}^2} + 2xy}}{{x + y}} \cr & = \frac{{{{\left( {2 + 2\sqrt 2 } \right)}^2} + 2 \times 2\sqrt 2 }}{{2 + 2\sqrt 2 }} \cr & = \frac{{4 + 8 + 8\sqrt 2 + 4\sqrt 2 }}{{2 + 2\sqrt 2 }} \cr & = \frac{{12 + 12\sqrt 2 }}{{2 + 2\sqrt 2 }} \cr & = \frac{{12\left( {1 + \sqrt 2 } \right)}}{{2\left( {1 + \sqrt 2 } \right)}} \cr & = 6 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 3 \cr & \Rightarrow {x^2} + 1 = 3x\,.....(i) \cr & \Rightarrow {\left( {{x^2} + 1} \right)^2} = {\left( {3x} \right)^2} \cr & \Rightarrow {x^4} + 1 + 2{x^2} = 9{x^2} \cr & \Rightarrow {x^4} + 1 = 7{x^2}\,.....(ii) \cr & \therefore \frac{{{x^4} + 3{x^3} + 5{x^2} + 3x + 1}}{{{x^4} + 1}} \cr & \Rightarrow \frac{{7{x^2} + 3{x^3} + 5{x^2} + 3x}}{{{x^4} + 1}} \cr & \Rightarrow \frac{{12{x^2} + 3{x^3} + 3x}}{{7{x^2}}} \cr & {\text{From equation (i)}} \cr & \Rightarrow \frac{{12x + 3\left( {{x^2} + 1} \right)}}{{7x}} \cr & \Rightarrow \frac{{12x + 3 \times 3x}}{{7x}} \cr & \Rightarrow \frac{{21x}}{{7x}} \cr & \Rightarrow 3 \cr} $$

$$\eqalign{ & x = \sqrt {a\root 3 \of {ab\sqrt {a\root 3 \of {ab} } } } ....... \propto \cr & {\text{Square both sides}} \cr & {x^2} = a\root 3 \of {abx} {\text{ }}\left( {\therefore \sqrt {a\root 3 \of {ab} } .... \propto } \right) \cr & {\text{Again cube both sides}} \cr & {x^6} = {a^3}abx \cr & {x^5} = {a^4}b \cr & x = \root 5 \of {{a^4}b} \cr} $$

$$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 1 \cr & {\text{Then, }}{\left( {x + \frac{1}{x}} \right)^2} = 1 + 2 \cr & \Rightarrow x + \frac{1}{x} = \sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} \cr & = {\left( {\sqrt 3 } \right)^3} - 3\sqrt 3 \cr & = 3\sqrt 3 - 3\sqrt 3 \cr & = 0 \cr & {\text{Then,}} \cr & {x^{102}} + {x^{96}} + {x^{90}} + {x^{84}} + {x^{78}} + {x^{72}} + 5 \cr} $$
  $$ = {x^{96}}\left( {{x^6} + 1} \right) + $$    $${x^{84}}\left( {{x^6} + 1} \right) + $$   $${x^{72}}\left( {{x^6} + 1} \right) + $$   $$5$$
  $$ = 5$$

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

$$\eqalign{ & a + b = 2c \cr & {\text{Taking }}a = 2,{\text{ }}b = 4{\text{ and }}c = 3 \cr & {\text{So,}}2 + 4 = 2 \times 3 \cr & \boxed{6 = 6} \cr & {\text{Now,}}\frac{a}{{a - c}} + {\text{ }}\frac{c}{{b - c}} \cr & = \frac{2}{{2 - 3}} + {\text{ }}\frac{3}{{4 - 3}} \cr & = \frac{2}{{ - 1}} + \frac{3}{1} \cr & = 1 \cr} $$

$$\eqalign{ & \frac{a}{b} = \frac{1}{2} \cr & {\text{Let }}a = x,{\text{ }}b = 2x \cr & {\text{Then,}}\frac{{\left( {2a - 5b} \right)}}{{\left( {5a + 3b} \right)}} \cr & = \frac{{2x - 10x}}{{5x + 6x}} \cr & {\text{ = }}\frac{{ - 8x}}{{11x}} \cr & = \frac{{ - 8}}{{11}} \cr} $$

$$\eqalign{ & \left( {\frac{3}{{15}}{a^5}{b^6}{c^3} \times \frac{5}{9}a{b^5}{c^4}} \right) \div \frac{{10}}{{27}}{a^2}b{c^3} \cr & \Rightarrow \frac{1}{9}{a^6}{b^{11}}{c^7} \div \frac{{10}}{{27}}{a^2}b{c^3} \cr & \Rightarrow \frac{{\frac{1}{9}{a^6}{b^{11}}{c^7}}}{{\frac{{10}}{{27}}{a^2}b{c^3}}} \cr & \Rightarrow \frac{3}{{10}}{a^4}{b^{10}}{c^4} \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & {\text{If }}x = 5 \cr & \therefore {x^2} - 2 + \frac{1}{{{x^2}}} \cr & = {\left( {x - \frac{1}{x}} \right)^2} \cr & = {\left( {5 - \frac{1}{5}} \right)^2} \cr & = {\left( {\frac{{24}}{5}} \right)^2} \cr & = \frac{{576}}{{25}} \cr} $$

$$\eqalign{ & \frac{x}{y}{\text{ = }}\frac{{a + 2}}{{a - 2}} \cr & \frac{{{x^2}}}{{{y^2}}}{\text{ = }}\frac{{{{\left( {a + 2} \right)}^2}}}{{{{\left( {a - 2} \right)}^2}}} \cr} $$
Applying componendo and dividendo
$$\eqalign{ & \therefore \frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} \cr & = \frac{{{{\left( {a + 2} \right)}^2} - {{\left( {a - 2} \right)}^2}}}{{{{\left( {a + 2} \right)}^2} + {{\left( {a - 2} \right)}^2}}} \cr & = \frac{{8a}}{{2{a^2} + 8}} \cr & = \frac{{4a}}{{{a^2} + 4}} \cr} $$

$$\eqalign{ & p{\left( {x + y} \right)^2} = 5{\text{ and }}q{\left( {x - y} \right)^2} = 3 \cr & {\text{Put the value of }}x = 2{\text{ and }}y = 1 \cr & p{\left( {2 + 1} \right)^2} = 5 \cr & \Leftrightarrow p = \frac{5}{9} \cr & q{\left( {2 - 1} \right)^2} = 3 \cr & \Leftrightarrow q = 3 \cr & {p^2}{\left( {x + y} \right)^2} + 4pqxy - {q^2}{\left( {x - y} \right)^2} \cr} $$
  $$ = {\left( {\frac{5}{9}} \right)^2}{\left( {2 + 1} \right)^2} + 4 \times \frac{5}{9} \times 3 \times $$      $$2 \times $$ $$1 - $$ $${\left( 3 \right)^2}$$ $${\left( {2 - 1} \right)^2}$$
$$\eqalign{ & = \frac{{25}}{{81}} \times 9 + \frac{{40}}{3} - 9 \cr & = \frac{{25}}{9} + \frac{{40}}{3} - 9 \cr & = \frac{{25 + 120 - 81}}{9} \cr & = \frac{{64}}{9} \cr & {\text{Put the value of p and q in option (A)}} \cr & {\text{Option 1}} \to 2\left( {p + q} \right) \cr & = 2\left( {\frac{5}{9} + 3} \right) \cr & = 2 \times \frac{{32}}{9} \cr & = \frac{{64}}{9} \cr & {\text{Option A is satisfied}} \cr & {\text{So, }}2\left( {p + q} \right){\text{ is answer}} \cr} $$

$$\eqalign{ & {\text{Given}} \cr & x + y + z = 6 \cr & xy + yz + zx = 10 \cr & {\text{To find }}{x^3} + {y^3} + {z^3} - 3xyz{\text{ = ?}} \cr & \Rightarrow {\text{ Using formula,}} \cr} $$
  $$ \Rightarrow {\left( {x + y + z} \right)^2} = $$   $${x^2} \,+$$ $$\, {y^2} \,+ $$ $$\, {z^2} \,+ $$ $$\, 2\left( {xy + yz + zx} \right)$$
$$\eqalign{ & \Rightarrow {6^2} = {x^2} + {y^2} + {z^2} + 2 \times 10 \cr & \Rightarrow 36 = {x^2} + {y^2} + {z^2} + 20 \cr & \Rightarrow {x^2} + {y^2} + {z^2} = 16 \cr} $$
  $$ \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = $$     $$\left( {x + y + z} \right)$$  $$\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right)$$
$$\eqalign{ & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = 6\left[ {16 - 10} \right] \cr & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = 6 \times 6 \cr & \Rightarrow {x^2} + {y^2} + {z^2} - 3xyz = 36 \cr} $$

$$\eqalign{ & {\text{Given ,}} \cr & \frac{{x + 1}}{{x - 1}} = \frac{a}{b} \cr & \left( {{\text{Using componendo & dividendo}}} \right) \cr & \Leftrightarrow \frac{x}{1} = \frac{{a + b}}{{a - b}} \cr & \Leftrightarrow x = \frac{{a + b}}{{a - b}}\,.....(i) \cr & {\text{Again,}}\frac{{1 - y}}{{1 + y}} = \frac{b}{a} \cr & \Leftrightarrow \frac{{1 + y}}{{1 - y}} = \frac{a}{b} \cr & \Leftrightarrow \frac{1}{y} = \frac{{a + b}}{{a - b}} \cr & \Leftrightarrow y = \frac{{a - b}}{{a + b}}\,.....(ii) \cr & {\text{From question,}} \cr & \frac{{x - y}}{{1 + xy}} \cr & \Rightarrow \frac{{\frac{{a + b}}{{a - b}} - \frac{{a - b}}{{a + b}}}}{{1 + \left( {\frac{{a + b}}{{a - b}}} \right)\left( {\frac{{a - b}}{{a + b}}} \right)}} \cr & \Rightarrow \frac{{{{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2}}}{{\left( {{a^2} - {b^2}} \right)\left( {1 + 1} \right)}} \cr & \Rightarrow \frac{{4ab}}{{2\left( {{a^2} - {b^2}} \right)}} \cr & \Rightarrow \frac{{2ab}}{{{a^2} - {b^2}}} \cr} $$

$$\eqalign{ & 999x + 888y = 1332 \cr & \underline {888x + 999y = 555} \cr & 1887\left( {x + y} \right) = 1887 \cr & \Rightarrow x + y = \frac{{1887}}{{1887}} \cr & \Rightarrow x + y = 1 \cr} $$

$$\eqalign{ & {\text{According to the question,}} \cr & {a^2} + {b^2} + {c^2} = ab + bc + ca \cr & {\text{Put }}a = 1 \cr & b = 1 \cr & c = 1 \cr & \therefore {a^2} + {b^2} + {c^2} = ab + bc + ca \cr & \Rightarrow {1^2} + {1^2} + {1^2} = 1 \times 1 + 1 \times 1 + 1 \times 1 \cr & \Rightarrow 1 + 1 + 1 = 1 + 1 + 1 \cr & \Rightarrow 3 = 3{\text{ }}\left( {{\text{Satisfy}}} \right) \cr & \therefore \frac{{a + c}}{b} \cr & = \frac{{1 + 1}}{1} \cr & = 2 \cr} $$