Answer: Option C. -> 123
$$\eqalign{
& x + \frac{1}{x} = 3 \cr
& \left( {{\text{Squaring both sides}}} \right) \cr
& {x^2} + \frac{1}{{{x^2}}} = 7 \cr
& {\text{On cubing both sides}} \cr
& {x^3} + \frac{1}{{{x^3}}} + 3.x.\frac{1}{x}\left( {x + \frac{1}{x}} \right) = 27 \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3 \times 3 = 27 \cr
& \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 18 \cr
& \therefore \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = 18 \times 7 \cr
& \Rightarrow \left( {{x^5} + \frac{1}{{{x^5}}}} \right) + \left( {x + \frac{1}{x}} \right) = 126 \cr
& \Rightarrow \left( {{x^5} + \frac{1}{{{x^5}}}} \right) + 3 = 126 \cr
& \Rightarrow \left( {{x^5} + \frac{1}{{{x^5}}}} \right) = 123 \cr} $$

Answer: Option B. -> 36
$$\eqalign{
& x - \frac{1}{x} = 3 \cr
& \left( {{\text{By cubing both sides}}} \right) \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3.x.\frac{1}{x}\left( {x - \frac{1}{x}} \right) = 27 \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3.\left( 3 \right) = 27 \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 27 + 9 \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 36 \cr} $$

Answer: Option B. -> (2, 1)
$$\eqalign{
& 2x + y = 5..............(i) \cr
& x + 2y = 4..............(ii) \cr
& {\text{Multiply equation (ii) by 2}} \cr
& 2x + 4y = 8............(iii) \cr} $$
Now subtracting equation (i) from (iii)
$$\eqalign{
& {\text{ }}2x + 4y = 8 \cr
& \mathop {}\limits_ - 2x\mathop + \limits_ - \,\,y\, = 5 \cr
& \overline {\underline {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3y = 3{\text{ }}} } \cr
& y = 1 \cr
& x = 2 \cr
& \therefore {\text{Intersection point = }}\left( {2,1} \right) \cr} $$

Answer: Option D. -> 3(x - 1)(y - 2)(z - 3)
$$\eqalign{
& x + y + z = 6 \cr
& {\left( {x - 1} \right)^3}{\text{ + }}{\left( {y - 2} \right)^3}{\text{ + }}{\left( {z - 3} \right)^3} \cr
& \therefore {\text{As, }}x + y + z = 6 \cr
& {\text{Take values}} \cr
& x = 1 \cr
& y = 2 \cr
& z = 3 \cr
& \left( {1 + 2 + 3} \right) = 6 \cr
& \therefore {\left( {1 - 1} \right)^3}{\text{ + }}{\left( {2 - 2} \right)^3}{\text{ + }}{\left( {3 - 3} \right)^3} \cr
& = 0 \cr} $$
Now assume values in options.
Option 'D' satisfies the given relation.
Hence 'D' is correct.

Answer: Option A. -> 1
$$\eqalign{
& a + b = 1 \cr
& {\text{By cubing,}} \cr
& {a^3} + {b^3} + 3ab\left( {a + b} \right) = {1^3} \cr
& \Rightarrow {a^3} + {b^3} + 3ab = 1\left[ {\because a + b = 1} \right] \cr
& \Rightarrow {a^3} + {b^3} + 3ab = k \cr
& {\text{From above both equations,}} \cr
& k = 1 \cr} $$

Answer: Option C. -> $$\frac{1}{3}$$
$$\eqalign{
& x + \frac{1}{x} = 99 \cr
& \therefore {x^2} + 1 = 99x \cr
& \Rightarrow 2\left( {{x^2} + 1} \right) = 2 \times 99x \cr
& \Rightarrow 2{x^2} + 2 = 198x \cr
& = \frac{{100x}}{{2{x^2} + 2 + 102x}} \cr
& = {\text{ }}\frac{{100x}}{{198x + 102x}} \cr
& = \frac{{100x}}{{300x}} \cr
& = \frac{1}{3} \cr} $$

Answer: Option C. -> 4
$$\eqalign{
& \frac{{4x - 3}}{x} + \frac{{4y - 3}}{y} + \frac{{4z - 3}}{z} = 0 \cr
& \Rightarrow \frac{{4x}}{x} - \frac{3}{x} + \frac{{4y}}{y} - \frac{3}{y} + \frac{{4z}}{z} - \frac{3}{z} = 0 \cr
& \Rightarrow 4 - \frac{3}{x} + 4 - \frac{3}{y} + 4 - \frac{3}{z} = 0 \cr
& \Rightarrow 12 - 3\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) = 0 \cr
& \Rightarrow - 3\left( {\frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right) = - 12 \cr
& \Rightarrow \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 4 \cr} $$

Answer: Option D. -> 8
$$\eqalign{
& xy = 8{\text{ }}\left( {{\text{Given}}} \right) \cr
& {\text{So, }}\left( {x,y} \right) = \left( {1,8} \right) \cr} $$
We have to question the options and check them.
$$\eqalign{
& \left( {8,1} \right) \cr
& \left( {2,4} \right) \cr
& \left( {4,2} \right) \cr
& \therefore {\text{ }}2x + y \cr
& = 2 \times 1 + 8 \cr
& = 10 \cr
& 2 \times 8 + 1 = 11 \cr
& 2 \times 2 + 4 = 8{\text{ }}\left( {{\text{Minimum}}} \right) \cr
& 2 \times 4 + 2 = 10 \cr} $$
Hence, in this question we have all the options.
So, take all the positive factor otherwise we should have to take - ve(negative) values also.
$$\eqalign{
& \left( {x,y} \right) = \left( {1,8} \right) \cr
& {\text{ }}\left( {8,1} \right) \cr
& {\text{ }}\left( {2,4} \right) \cr
& {\text{ }}\left( {4,2} \right) \cr
& {\text{ }}\left( { - 1, - 8} \right) \cr
& {\text{ }}\left( { - 8, - 1} \right) \cr
& {\text{ }}\left( { - 2, - 4} \right) \cr
& {\text{ }}\left( { - 4, - 2} \right) \cr} $$

Answer: Option B. -> 30
$$\eqalign{
& {a^2} - 4a - 1 = 0 \cr
& {a^2} - 1 = 4a \cr
& a - \frac{1}{a} = 4 \cr
& {\text{Squring both sides}} \cr
& {a^2} + \frac{1}{{{a^2}}} - 2 = 16 \cr
& \Rightarrow {a^2} + \frac{1}{{{a^2}}} = 18 \cr
& \therefore {a^2} + \frac{1}{{{a^2}}} + 3a - \frac{3}{a} \cr
& = {a^2} + \frac{1}{{{a^2}}} + 3\left( {a - \frac{1}{a}} \right) \cr
& = 18 + 3 \times 4 \cr
& = 18 + 12 \cr
& = 30 \cr} $$

Answer: Option C. -> $$\frac{{2abc}}{{bc + ac - ab}}$$
$$\eqalign{
& \frac{{xy}}{{x + y}} = a,\,\frac{{xz}}{{x + z}} = b,\,\frac{{yz}}{{y + z}} = c{\text{ }} \cr
& {\text{Now,}} \cr
& \frac{{x + y}}{{xy}} = \frac{1}{a} \cr
& \frac{{x + z}}{{xz}} = \frac{1}{b} \cr
& \frac{{y + z}}{{yz}} = \frac{1}{c} \cr
& \Rightarrow \frac{1}{y} + \frac{1}{x} = \frac{1}{a},\frac{1}{z} + \frac{1}{x} = \frac{1}{b},\frac{1}{x} + \frac{1}{y} = \frac{1}{c} \cr
& {\text{Now we have to find the value of }}x \cr
& \therefore \frac{1}{a} + \frac{1}{b} - \frac{1}{c} = \frac{1}{y} + \frac{1}{x} + \frac{1}{z} + \frac{1}{x} - \frac{1}{y} - \frac{1}{z} \cr
& \therefore \frac{1}{a} + \frac{1}{b} - \frac{1}{c} = \frac{2}{x} \cr
& \Rightarrow \frac{{bc + ac - ab}}{{abc}} = \frac{2}{x} \cr
& \Rightarrow x = \frac{{2abc}}{{bc + ac - ab}} \cr} $$