Answer: Option C. -> $$\frac{1}{2}$$
$$\eqalign{
& x + \frac{1}{x} = 5{\text{ then, }}\frac{{5x}}{{{x^2} + 5x + 1}} \cr
& \Rightarrow \frac{5}{{x + 5 + \frac{1}{x}}} \cr
& \Rightarrow \frac{5}{{x + \frac{1}{x} + 5}} \cr
& \Rightarrow \frac{5}{{5 + 5}} \cr
& \Rightarrow \frac{1}{2} \cr} $$

Answer: Option B. -> 18
$$\eqalign{
& \frac{1}{a}\left( {{a^2} + 1} \right) = 3 \cr
& \Rightarrow a + \frac{1}{a} = 3 \cr
& \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = {3^3} \cr
& \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\left( 3 \right) = {3^3} \cr
& \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 27 - 9 \cr
& \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 18 \cr
& \Rightarrow \frac{{{a^6} + 1}}{{{a^3}}} = 18 \cr} $$

Answer: Option C. -> -2
$$\eqalign{
& {\left( {x - 5} \right)^2} + {\left( {y - 2} \right)^2} + {\left( {z - 9} \right)^2} = 0 \cr
& {\text{It is possible only when }} \cr
& x - 5 = 0 \cr
& x = 5 \cr
& y - 2 = 0 \cr
& y = 2 \cr
& z - 9 = 0 \cr
& z = 9 \cr
& \therefore x + y - z \cr
& = 5 + 2 - 9 \cr
& = 7 - 9 \cr
& = - 2 \cr} $$

Answer: Option D. -> 9
$$\eqalign{
& \therefore {a^3} + {b^3} + {c^3} - 3abc \cr
& = \frac{1}{2}\left( {a + b + c} \right)\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right] \cr
& \therefore \frac{{{x^3} + {y^3} + {z^3} - 3xyz}}{{x - y + z}} \cr} $$
$$ = \frac{{\frac{1}{2}\left( {x + y + z} \right)\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right]}}{{x - y + z}}$$
$$\eqalign{
& = \frac{{\frac{1}{2}\left( {999 + 1000 + 1001} \right)\left( {1 + 1 + 4} \right)}}{{999 - 1000 + 1001}} \cr
& = \frac{{\frac{1}{2} \times 6 \times 3000}}{{1000}} \cr
& = 9 \cr} $$

Answer: Option D. -> 4
$$\eqalign{
& \frac{a}{{1 - 2a}} + \frac{b}{{1 - 2b}} + \frac{c}{{1 - 2c}} = \frac{1}{2} \cr
& {\text{Multiply by 2 both side}} \cr
& \Rightarrow \frac{{2a}}{{1 - 2a}} + \frac{{2b}}{{1 - 2b}} + \frac{{2c}}{{1 - 2c}} = 1 \cr
& {\text{Adding 3 both side}} \cr} $$
$$ \Rightarrow 1 + \frac{{2a}}{{1 - 2a}} + 1 + \frac{{2b}}{{1 - 2b}} + 1 + $$       $$\frac{{2c}}{{1 - 2c}} = $$  $$1 + 3$$
$$ \Rightarrow \frac{1}{{1 - 2a}} + \frac{1}{{1 - 2b}} + \frac{1}{{1 - 2c}} = 4$$

Answer: Option B. -> $$\frac{1}{3}$$
$$\eqalign{
& \frac{{5x}}{{4{x^2} + 10x + 1}} \cr
& = \frac{5x}{{x\left( {4x + 10 + \frac{1}{x}} \right)}} \cr
& = \frac{5}{{4x + \frac{1}{x} + 10}} \cr
& = \frac{5}{{5 + 10}} \cr
& = \frac{5}{{15}} \cr
& = \frac{1}{3} \cr} $$

Answer: Option C. -> 3abc
$$\eqalign{
& a + b + c = 0 \cr
& {\text{Let, }}{a^3} + {b^3} + {c^3} = T \cr} $$
$$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \frac{1}{2}\left( {a + b + c} \right)$$      $$\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right]$$
$$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( 0 \right)$$     $$\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right]$$
$$\eqalign{
& \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 0 \cr
& \Rightarrow {a^3} + {b^3} + {c^3} = 3abc \cr} $$

Answer: Option C. -> 64
$$\eqalign{
& {\text{We know that }} \cr
& {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) \cr
& {x^3} - p = \left( {x - 4} \right)\left( {{x^2} + 4x + 16} \right) \cr
& \Rightarrow {x^3} - p = \left( {{x^3} - {4^3}} \right) \cr
& \Rightarrow p = {4^3}{\text{ }}\left( {{\text{By comparison}}} \right) \cr
& {\text{So, }}p = 64 \cr} $$

Answer: Option B. -> 4690
$${x^3} + {y^3} + {z^3} - 3xyz$$
$$ = \frac{1}{2}\left( {x + y + z} \right)$$  $$\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right]$$
$$ = \frac{1}{2}\left( {222 + 223 + 225} \right)$$    $$\left[ {{{\left( {222 - 223} \right)}^2} + {{\left( {223 - 225} \right)}^2} + {{\left( {225 - 222} \right)}^2}} \right]$$
$$\eqalign{
& = \frac{1}{2}\left( {670} \right)\left( {1 + 4 + 9} \right) \cr
& = \frac{1}{2} \times 670 \times 14 \cr
& = 4690 \cr} $$

Answer: Option A. -> 0
$$\eqalign{
& c + \frac{1}{c} = \sqrt 3 \cr
& {\text{On cubing both side}} \cr
& \Rightarrow {\left( {c + \frac{1}{c}} \right)^3} = 3\sqrt 3 \cr
& \Rightarrow {c^3} + \frac{1}{{{c^3}}} + 3.c.\frac{1}{c}\left( {c + \frac{1}{c}} \right) = 3\sqrt 3 \cr
& \Rightarrow {c^3} + \frac{1}{{{c^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr
& \Rightarrow {c^3} + \frac{1}{{{c^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr
& \Rightarrow {c^3} + \frac{1}{{{c^3}}} = 0 \cr} $$