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

Answer: Option C. -> 7
$$\eqalign{
& {p^2} + {q^2} = 7pq \cr
& \frac{p}{q}{\text{ + }}\frac{q}{p} = 7 \cr} $$
(Divide whole equation by pq)

Answer: Option A. -> 8
$$\eqalign{
& x = {3^{\frac{1}{3}}} - {3^{ - \frac{1}{3}}} \cr
& \Rightarrow x + {3^{ - \frac{1}{3}}} = {3^{\frac{1}{3}}}\,\,\left( {{\text{cubing both sides}}} \right) \cr
& \Rightarrow {x^3} + \frac{1}{3} + 3 \times x \times {3^{ - \frac{1}{3}}}\left( {x + {3^{ - \frac{1}{3}}}} \right) = 3 \cr
& \Rightarrow {x^3} + \frac{1}{3} + 3 \times x \times {3^{ - \frac{1}{3}}} \times {3^{\frac{1}{3}}} = 3 \cr
& \,\,\,\,\,\,\,\,\,\left( {{\text{multiply both sides by 3}}} \right) \cr
& \Rightarrow 3{x^3} + 9x = 9 - 1 \cr
& \Rightarrow 3{x^3} + 9x = 8 \cr} $$

Answer: Option D. -> 15
$$\eqalign{
& {\text{Give,}} \cr
& p + \frac{1}{{p + 2}} = 1 \cr
& {\text{Adding 2 both sides}} \cr
& p + 2 + \frac{1}{{p + 2}} = 1 + 2 = 3 \cr
& {\text{Let}}\left( {p + 2} \right) = a\,\,\&\,\, \frac{1}{{p + 2}} = b \cr
& a + b = 3 \cr
& {\text{Cubbing both sides}} \cr
& {\left( {a + b} \right)^3} = {3^3} \cr
& {a^3} + {b^3} + 3ab\left( {a + b} \right) = 27 \cr
& {a^3} + {b^3} + 3 \times ab \times 3 = 27 \cr
& {a^3} + {b^3} + 9ab = 27......\left( {\text{i}} \right) \cr
& {\text{Now,}} \cr
& a \times b = \left( {p + 2} \right) \times \frac{1}{{\left( {p + 2} \right)}} \cr
& a \times b = 1........\left( {{\text{ii}}} \right) \cr
& {\text{Put the a & b value in equation}}\left( {\text{i}} \right) \cr
& {a^3} + {b^3} + 9 \times 1 = 27 \cr
& {a^3} + {b^3} = 27 - 9 = 18 \cr
& \therefore {\left( {p + 2} \right)^3} + \frac{1}{{{{\left( {p + 2} \right)}^3}}} - 3 \cr
& = {a^3} + {b^3} - 3 \cr
& = 18 - 3 \cr
& = 15 \cr} $$

Answer: Option C. -> $$\frac{{{m^3}}}{{{x^3}{y^3}}}$$
$$\eqalign{
& xy\left( {x + y} \right) = m \cr
& {\text{find }}{x^3} + {y^3} + 3m \cr
& xy\left( {x + y} \right) = m\, . . . . . (i) \cr
& \left( {x + y} \right) = \frac{m}{xy} \cr
&{\text{Cubing both side}} \cr
& \Rightarrow {x^3} + {y^3} + 3.xy\left( {x + y} \right) = \frac{{{m^3}}}{{{x^3}{y^3}}} \cr
& {\text{From equation (i)}} \cr
& \Rightarrow {x^3} + {y^3} + 3.m = \frac{{{m^3}}}{{{x^3}{y^3}}} \cr
& \Rightarrow {x^3} + {y^3} + 3m = \frac{{{m^3}}}{{{x^3}{y^3}}} \cr} $$

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

Answer: Option B. -> 2
$$\eqalign{
& x + \frac{1}{x} = 2{\text{ }} \cr
& {\text{But }}x = 1 \cr
& {\text{So, }}{x^2} + \frac{1}{{{x^2}}} = {1^2} + \frac{1}{{{1^2}}} = 2 \cr} $$

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

Answer: Option C. -> 3
$$\eqalign{
& \Rightarrow {p^3} - {q^3} = \left( {p - q} \right)\left\{ {{{\left( {p - q} \right)}^2} + xpq} \right\} \cr
& \Rightarrow \left( {p - q} \right)\left\{ {{p^2} + {q^2} + pq} \right\} = \left( {p - q} \right)\left\{ {{p^2} + {q^2} - 2pq + xpq} \right\} \cr
& \Rightarrow {p^2} + {q^2} + pq = {p^2} + {q^2} - 2pq + xpq \cr
& \Rightarrow 3pq = xpq \cr
& \Rightarrow x = 3 \cr} $$

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