$$\eqalign{ & a + b + c = 9 \cr & {\text{For minimum value}} \cr & a = b = c \cr & \Rightarrow 3a = 9 \cr & \Rightarrow a = \frac{9}{3} \cr & \Rightarrow a = 3 \cr & {\text{For minimum value}} \cr & a = b = c = 3 \cr & \therefore {a^2} + {b^2} + {c^2} \cr & \Rightarrow {3^2} + {3^2} + {3^2} \cr & \Rightarrow 9 + 9 + 9 \cr & \Rightarrow 27 \cr} $$

$$\eqalign{ & {a^2} + {b^2} + 4{c^2} = 2\left( {a + b - 2c} \right) - 3 \cr & \Rightarrow {a^2} + {b^2} + 4{c^2} - 2a - 2b + 4c + 3 = 0 \cr & \Rightarrow {a^2} - 2a + 1 + {b^2} - 2b + 1 + 4{c^2} + 4c + 1 = 0 \cr & \Rightarrow {\left( {a - 1} \right)^2} + {\left( {b - 1} \right)^2} + {\left( {2c + 1} \right)^2} = 0 \cr & \cr & \therefore a - 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = 1 \cr & \,\,\,\,\,\,\,b - 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b = 1 \cr & \,\,\,\,2c + 1 = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c = \frac{{ - 1}}{2} \cr & \cr & \therefore {a^2} + {b^2} + {c^2} \cr & \Rightarrow 1 + 1 + \frac{1}{4} \cr & \Rightarrow 2 + \frac{1}{4} \cr & \Rightarrow \frac{9}{4} \cr & \Rightarrow 2\frac{1}{4} \cr} $$

$$\eqalign{ & \left( {x - a} \right)\left( {x - b} \right) = 1{\text{ }} \cr & \Rightarrow \left( {x - a} \right) = \frac{1}{{\left( {x - b} \right)}} \cr & {\text{ }}a - b + 5 = 0 \cr & \Rightarrow {\text{ }}a - b = - 5{\text{ }}\left( {{\text{Given}}} \right) \cr & {\text{Add and subtract }}x \cr & \Rightarrow a - b + x - x = - 5 \cr & \Rightarrow \left( {a - x} \right) + \left( {x - b} \right) = - 5 \cr & \Rightarrow \left( {x - b} \right) - \left( {x - a} \right) = - 5 \cr & \Rightarrow \left( {x - a} \right) - \left( {x - b} \right) = + 5 \cr & \Rightarrow \left( {x - a} \right) - \frac{1}{{\left( {x - a} \right)}} = + 5 \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\left( {x - a} \right)^3} - \frac{1}{{{{\left( {x - a} \right)}^3}}} - 3\left( {x - a} \right)\frac{1}{{\left( {x - a} \right)}}\left( {\left( {x - a} \right) - \frac{1}{{\left( {x - a} \right)}}} \right) = {\left( 5 \right)^3} \cr & \Rightarrow {\left( {x - a} \right)^3} - \frac{1}{{{{\left( {x - a} \right)}^3}}} - 3 \times 5 = 125 \cr & \Rightarrow {\left( {x - a} \right)^3} - \frac{1}{{{{\left( {x - a} \right)}^3}}} = 125 + 15 \cr & \Rightarrow {\left( {x - a} \right)^3} - \frac{1}{{{{\left( {x - a} \right)}^3}}} = 140 \cr} $$

$$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & \Rightarrow {x^2} + 1 = 3x \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & {\text{Squaring both sides}} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 9 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 7 \cr & \therefore {x^2} + x + \frac{1}{x} + \frac{1}{{{x^2}}} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + x + \frac{1}{x} \cr & \Rightarrow 7 + 3 \cr & \Rightarrow 10 \cr} $$

$$\eqalign{ & {a^2} + {b^2} = 5ab \cr & \Rightarrow \frac{{{a^2}}}{{ab}} + \frac{{{b^2}}}{{ab}} = 5 \cr & \Rightarrow \frac{a}{b}{\text{ + }}\frac{b}{a}{\text{ = 5}} \cr & {\text{Squaring the both sides}} \cr & \Rightarrow {\left( {\frac{a}{b}} \right)^2}{\text{ + }}{\left( {\frac{b}{a}} \right)^2} + 2 \times \frac{a}{b} \times \frac{b}{a} = 25 \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}}{\text{ + }}\frac{{{b^2}}}{{{a^2}}} = 25 - 2 \cr & \Rightarrow \frac{{{a^2}}}{{{b^2}}}{\text{ + }}\frac{{{b^2}}}{{{a^2}}} = 23 \cr} $$

$$\eqalign{ & x = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ and y}} = \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \therefore x = \frac{1}{y} \cr & \Leftrightarrow xy = 1 \cr & x + y = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}}{\text{ + }}\frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & \Rightarrow x + y = \frac{{5 + 1 + 2\sqrt 5 + 5 + 1 - 2\sqrt 5 }}{{5 - 1}} \cr & \Rightarrow x + y = \frac{{12}}{4} \cr & \Rightarrow x + y = 3 \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = {\left( 3 \right)^2} - 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} = 7 \cr & {\text{Now,}}\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} \cr & = \frac{{{x^2} + {y^2} + xy}}{{{x^2} + {y^2} - xy}} \cr & = \frac{{7 + 1}}{{7 - 1}} \cr & = \frac{8}{6} \cr & = \frac{4}{3} \cr} $$

$$\eqalign{ & {x^4} + 2{x^3} + a{x^2} + bx + 9 \cr & {\text{Put }}x = 1 \cr & = 1 + 2 \times 1 + a + b + 9 \cr & = 1 + 2 + a + b + 9 \cr & = 13 + a + b \cr} $$
To make a perfect square numbers value of a + b must be either 3 or 13
Now, option (B) a = 6, b = 7
$$\eqalign{ & \therefore a + b = 13 \cr & {\text{make perfect square}} \cr & \left( {25 = {5^2}} \right) \cr} $$

$$\eqalign{ & {x^2} + \frac{1}{5}x + {a^2} \cr & {{\text{A}}^2} + {\text{2}} \times {\text{AB}} + {{\text{B}}^2} = {\left( {{\text{A}} + {\text{B}}} \right)^2} \cr & {x^2} + 2 \times \frac{1}{{10}} \times x + {a^2} = {\left( {x + \frac{1}{{10}}} \right)^2} \cr & {\text{A}} = x \cr & {\text{B}} = \frac{1}{{10}} \cr & {\text{B}} = a = \frac{1}{{10}} \cr} $$

$$\eqalign{ & \therefore {\text{When}}\left( {x - 1} \right) = 0 \cr & x = 1 \cr & {x^2} + {k_1}x + {k_2}{\text{ }} = 0 \cr & \Rightarrow 1 + {k_1} + {k_2}{\text{ }} = 0 \cr & \Rightarrow {k_1} + {k_2} = - 1\,.......(i) \cr & {\text{When}}\left( {x + 3} \right) = 0 \cr & x = - 3 \cr & 9 - 3{k_1} + {k_2}{\text{ }} = 0 \cr & \Rightarrow - 3{k_1} + {k_2}{\text{ }} = - 9\,.......(ii) \cr & {\text{From equation (i) and (ii)}} \cr & {k_1} = 2,\,\,\,\,\,\,{\text{ }}{k_2} = - 3 \cr} $$

$$\frac{{x - {a^2}}}{{b + c}} + \frac{{x - {b^2}}}{{c + a}} + \frac{{x - {c^2}}}{{a + b}} = 4\left( {a + b + c} \right)$$
Note : In such type of question to save your valuable time assume values as per your need which make your calculation easier.
Assume a = 1, b = 0, c = 1
$$\eqalign{ & {\text{Make sure there will be no }}\left( {\frac{0}{0}} \right){\text{ form}} \cr & \therefore \frac{{x - 1}}{{1 + 0}} + \frac{{x - 0}}{{1 + 1}} + \frac{{x - 1}}{{1 + 0}} = 4 \cr & \Rightarrow x - 1 + \frac{x}{2} + x - 1 = 4 \times 2 \cr & \Rightarrow x + \frac{x}{2} + x = 8 + 2 \cr & \Rightarrow \frac{{5x}}{2} = 10 \cr & \Rightarrow x = 4 \cr & {\text{Now put values in options take option}} \cr & \left( \text{A} \right),{\left( {a + b + c} \right)^2} = {\left( {1 + 0 + 1} \right)^2} = 4 \cr} $$

$$\eqalign{ & \frac{a}{b}{\text{ = }}\frac{4}{5}{\text{ and }}\frac{b}{c}{\text{ = }}\frac{{15}}{{16}} \cr & \Rightarrow \frac{a}{b} \times \frac{b}{c}{\text{ = }}\frac{4}{5} \times \frac{{15}}{{16}} \cr & \Rightarrow \frac{a}{c} = \frac{3}{4} \cr & \therefore \frac{{18{c^2} - 7{a^2}}}{{45{c^2} + 20{a^2}}} \cr & = \frac{{{c^2}\left( {18 - 7\frac{{{a^2}}}{{{c^2}}}} \right)}}{{{c^2}\left( {45 + 20\frac{{{a^2}}}{{{c^2}}}} \right)}} \cr & = \frac{{18 - 7{{\left( {\frac{a}{c}} \right)}^2}}}{{45 + 20{{\left( {\frac{a}{c}} \right)}^2}}} \cr & = \frac{{18 - 7 \times \frac{9}{{16}}}}{{45 + 20 \times \frac{9}{{16}}}} \cr & = \frac{{18 - \frac{{63}}{{16}}}}{{45 + \frac{{45}}{4}}} \cr & = \frac{{225 \times 4}}{{16 \times 225}} \cr & = \frac{1}{4} \cr} $$

$$\eqalign{ & {x^3} + \frac{1}{{{x^3}}} = 0{\text{ }} \cr & {\text{It is possible only when}} \cr & x + \frac{1}{x} = \sqrt 3 {\text{ }} \cr & {\text{So, }}{\left( {x + \frac{1}{x}} \right)^4} = {\left( {\sqrt 3 } \right)^4} \cr & \Leftrightarrow {\left( {x + \frac{1}{x}} \right)^4} = 9 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 2{\text{ }} \cr & {\text{But }}x = 1 \cr & {\text{So, }}{x^2} + \frac{1}{{{x^2}}} = {1^2} + \frac{1}{{{1^2}}} = 2 \cr} $$

$$\eqalign{ & {\left( {{\text{ }}a + \frac{1}{a}} \right)^2} = 3 \cr & \Rightarrow a + \frac{1}{a} = \sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr & \Rightarrow {a^6} = - 1 \cr & \Rightarrow \left( {{a^6} + 1} \right) = 0 \cr & {\text{So,}}{a^{18}} + {\text{ }}{a^{12}} + {\text{ }}{a^6} + {\text{ 1}} \cr & \Rightarrow {a^{12}}\left( {{a^6} + 1} \right) + {\text{ }}{a^6} + {\text{ 1}} \cr & \Rightarrow {a^{12}}\left( 0 \right) + 0 \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & xy + yz + zx = 0 \cr & \therefore xy + zx = - yz \cr & \Rightarrow xy + yz = - zx \cr & \Rightarrow yz + zx = - xy \cr & \therefore \frac{1}{{{x^2} - yz}} + \frac{1}{{{y^2} - zx}} + \frac{1}{{{z^2} - xy}} \cr} $$
Putting values of -yz, -zx, -xy from above
$$ \Rightarrow \frac{1}{{{x^2} + \left( {xy + zx} \right)}} + \frac{1}{{{y^2} + \left( {xy + yz} \right)}}$$       $$ + \frac{1}{{{z^2} + \left( {yz + zx} \right)}}$$
$$ \Rightarrow \frac{1}{{x\left( {x + y + z} \right)}} + \frac{1}{{y\left( {x + y + z} \right)}}$$       $$ + \frac{1}{{z\left( {x + y + z} \right)}}$$
$$\eqalign{ & \Rightarrow \frac{1}{{\left( {x + y + z} \right)}}\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) \cr & \Rightarrow \frac{1}{{\left( {x + y + z} \right)}}\left( {\frac{{zy + xz + xy}}{{xyz}}} \right) \cr & \Rightarrow \frac{1}{{x + y + z}} \times 0 \cr & \Rightarrow 0 \cr} $$

According to the question,
Divisor = 2 × remainder
= 2 × 80
= 160
Again, 4 × quotient = 160
⇒ Quotient = $$\frac{{160}}{4}$$ = 40
∴ x = Divisor × Quotient + Remainder
x = 160 × 40 + 80 = 6480

$$\eqalign{ & {a^2} + {b^2} + {c^2} = 16,{\text{ }}{x^2} + {y^2} + {z^2} = 25{\text{ }} \cr & {\text{But }}b = c = 0,{\text{ But }}y = z = 0 \cr & {\text{}}a = 4{\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 5 \cr & {\text{Now, }} \cr & ax + by + cz = 20 \cr & 4 \times 5 + 0 + 0 = 20 \cr & 20 = 20{\text{ }}\left( {{\text{Satisfy}}} \right) \cr & {\text{Now, }}\frac{{a + b + c}}{{x + y + z}} \cr & = \frac{{4 + 0 + 0}}{{5 + 0 + 0}} \cr & = \frac{4}{5} \cr} $$

$$\eqalign{ & \frac{{5x}}{{2{x^2} + 5x + 1}} = \frac{1}{3} \cr & \Rightarrow \frac{5}{{\frac{{2{x^2}}}{x} + \frac{{5x}}{x} + \frac{1}{x}}} = \frac{1}{3} \cr & \Rightarrow \frac{5}{{2x + \frac{1}{x} + 5}} = \frac{1}{3} \cr & \Rightarrow 2x + \frac{1}{x} + 5 = 15 \cr & \Rightarrow 2x + \frac{1}{x} = 10 \cr & {\text{Divide by 2 both sides }} \cr & \Rightarrow x + \frac{1}{{2x}} = \frac{{10}}{2} \cr & \Rightarrow 2x + \frac{1}{x} = 5 \cr} $$

$$\frac{1}{{{x^2}}} + \frac{1}{{{y^2}}} + \frac{1}{{{z^2}}} = \frac{1}{{xy}} + \frac{1}{{yz}} + \frac{1}{{zx}}$$
Go through option D take x = y = z
$$\eqalign{ & \frac{1}{{{x^2}}} + \frac{1}{{{x^2}}} + \frac{1}{{{x^2}}} = \frac{1}{{x^2}} + \frac{1}{{x^2}} + \frac{1}{{x^2}} \cr & \therefore {\text{Option 'D' is right}} \cr} $$

$$\eqalign{ & a = 2,{\text{ }}b = - 3 \cr & {\text{So,}}27{a^3} - 54{a^2}b + 36a{b^2} - 8{b^3} \cr} $$
$$ = 27 \times {\left( 2 \right)^3} - 54 \times {\left( 2 \right)^2} \times \left( { - 3} \right) + $$       $$36 \times $$ $$\left( 2 \right) \times $$ $${\left( { - 3} \right)^2}$$ $$ -\, 8 \times $$ $${\left( { - 3} \right)^3}$$
$$\eqalign{ & = 27 \times 8 - 54 \times 4 \times - 3 + 36 \times 2 \times 9 - 8 \times - 27 \cr & = 216 + 216 + 648 + 648 \cr & = 1728 \cr} $$

$$\eqalign{ & {\text{But }}a = 1 \cr & {a^3} + \frac{1}{{{a^3}}} = 2 \cr & \Rightarrow {1^3} + \frac{1}{{{1^3}}} = 2 \cr & \Rightarrow 2 = 2{\text{ }}\left( {{\text{Satisfy}}} \right) \cr & {\text{So, }}\frac{{{a^2} + 1}}{a} \cr & = \frac{{{1^2} + 1}}{1} \cr & = 2 \cr} $$

$$\eqalign{ & {\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy \cr & \Rightarrow {\left( {x + y} \right)^2} = 29 + 2 \times 10 \cr & \Rightarrow {\left( {x + y} \right)^2} = 49 \cr & \therefore x + y = 7 \cr & {\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy \cr & \Rightarrow {\left( {x - y} \right)^2} = 29 - 2 \times 10 \cr & \Rightarrow {\left( {x - y} \right)^2} = 9 \cr & \therefore \left( {x - y} \right) = 3 \cr & \therefore \frac{{x + y}}{{x - y}} = \frac{7}{3} \cr} $$

$$\eqalign{ & {p^2} + {q^2} = 7pq \cr & \frac{p}{q}{\text{ + }}\frac{q}{p} = 7 \cr} $$
(Divide whole equation by pq)

$$\eqalign{ & 2x + \frac{2}{{9x}} = 4 \cr & {\text{Multiply by }}\frac{3}{2}{\text{ on both sides}} \cr & \Rightarrow 3x + \frac{1}{{3x}} = 6 \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\left( {3x + \frac{1}{{3x}}} \right)^3} = {6^3} \cr & \Rightarrow 27{x^3} + \frac{1}{{27{x^3}}} + 3 \times 3x \times \frac{1}{{3x}}\left( {3x + \frac{1}{{3x}}} \right) = 216 \cr & \Rightarrow 27{x^3} + \frac{1}{{27{x^3}}} + 3 \times 6 = 216 \cr & \Rightarrow 27{x^3} + \frac{1}{{27{x^3}}} = 216 - 18 \cr & \Rightarrow 27{x^3} + \frac{1}{{27{x^3}}} = 198 \cr} $$

$$\eqalign{ & {x^2} + \frac{{{\text{ }}1}}{{{x^2}}} = 2 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} + 2.x.\frac{1}{x} = 2 \cr & \Rightarrow x - \frac{1}{x} = 2 - 2 \cr & \Rightarrow x - \frac{1}{x} = 0 \cr} $$

$$\eqalign{ & x + y = \sqrt 3 \,.......{\text{(i)}} \cr & x - y = \sqrt 2 \,.......(ii) \cr & {\text{From equation (i) and (ii)}} \cr & x = \frac{{\sqrt 3 + \sqrt 2 }}{2} \cr & y = \frac{{\sqrt 3 - \sqrt 2 }}{2} \cr & {\text{So, }}8xy\left( {{x^2} + {y^2}} \right) \cr & = 8 \times \frac{{\sqrt 3 + \sqrt 2 }}{2} \times \frac{{\sqrt 3 - \sqrt 2 }}{2}\left[ {\frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{4} + \frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{4}} \right] \cr & = 2\left( {3 - 2} \right)\left[ {\frac{{3 + 2 + 2\sqrt 6 + 3 + 2 - 2\sqrt 6 }}{4}} \right] \cr & = 2 \times 1 \times \frac{{10}}{4} \cr & = 5 \cr} $$

$$\frac{{\left( {x + \frac{1}{x}} \right)}}{{\left( {x - \frac{1}{x}} \right)}} = \frac{5}{3}$$
By Componendo and Dividendo
$$\eqalign{ & \Rightarrow \frac{{x + \frac{1}{x} + x - \frac{1}{x}}}{{x + \frac{1}{x} - x + \frac{1}{x}}} = \frac{{5 + 3}}{{5 - 3}} \cr & \Rightarrow \frac{{2x}}{{\frac{2}{x}}} = \frac{8}{2} \cr & \Rightarrow {x^2} = 4 \cr & \Rightarrow x = \pm 2 \cr} $$

$$\eqalign{ & \because p = 99 \cr & p\left( {{p^2} + 3p + 3} \right) \cr & = {p^3} + 3{p^2} + 3p + 1 - 1 \cr & = {\left( {p + 1} \right)^3} - 1 \cr & = {\left( {99 + 1} \right)^3} - 1 \cr & = {\left( {100} \right)^3} - 1 \cr & = 1000000 - 1 \cr & = 999999 \cr} $$

$$\eqalign{ & \frac{1}{p} + \frac{1}{q} = \frac{1}{{p + q}} \cr & \Rightarrow \frac{{p + q}}{{pq}} = \frac{1}{{p + q}} \cr & \Rightarrow {\left( {p + q} \right)^2} = pq \cr & \Rightarrow \left( {{p^2} + {q^2} + 2pq - pq} \right) = 0 \cr & \Rightarrow \left( {{p^2} + {q^2} + pq} \right) = 0 \cr & {\text{Multiply by }}\left( {p - q} \right){\text{ both side}} \cr & \Rightarrow \left( {p - q} \right)\left( {{p^2} + {q^2} + pq} \right) = \left( {p - q} \right) \times 0 \cr & \Rightarrow {p^3} - {q^3} = 0 \cr} $$

$$\eqalign{ & \frac{{{{\left( {a + b} \right)}^2}}}{{{{\left( {a - b} \right)}^2}}} = \frac{{25}}{4} \cr & \Rightarrow \frac{{a + b}}{{a - b}} = \frac{5}{2} \cr & \Rightarrow {\text{By Componendo & Dividendo}} \cr & \Rightarrow \frac{{a + b + a - b}}{{a + b - a + b}} = \frac{{5 + 2}}{{5 - 2}} \cr & \Rightarrow \frac{{2a}}{{2b}} = \frac{7}{3} \cr & \Rightarrow \frac{a}{b} = \frac{7}{3} \cr & {\text{Now, the value of}} \cr & \Rightarrow {a^2} + {b^2} + 3ab \cr & \Rightarrow {7^2} + {3^2} + 3.7.3 \cr & \Rightarrow 49 + 9 + 63 \cr & \Rightarrow 121 \cr} $$