Answer: Option A. -> $$\frac{{20}}{{101}}$$
$$\left[ {\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\, - \,\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,} \right]$$              $$ ÷ $$  $$\left[ {\,\left( {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,\, + \,\,\left( {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right)\,} \right]$$
$$ {\text{Let }}\left[ {1 + \frac{1}{{10 + \frac{1}{{10}}}}} \right] = a,$$    $${\text{ }}\left[ {1 - \frac{1}{{10 + \frac{1}{{10}}}}} \right] = b $$
$$\eqalign{
& \Rightarrow \left( {{a^2} - {b^2}} \right) \div \left( {a + b} \right) = a - b = ? \cr
& \Rightarrow a = 1 + \frac{{10}}{{101}} \cr
& \Rightarrow a = \frac{{111}}{{101}} \cr
& \Rightarrow b = 1 - \frac{{10}}{{101}} \cr
& \Rightarrow b = \frac{{91}}{{101}} \cr
& \Rightarrow a - b = \frac{{111}}{{101}} - \frac{{91}}{{101}} \cr
& \Rightarrow a - b = \frac{{20}}{{101}} \cr} $$

Answer: Option D. -> $$\frac{{63}}{{61}}$$
$$\eqalign{
& {\text{Given,}} \cr
& x = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}{\text{ , }}y = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr
& {\text{Find, }}\frac{{{x^2} + xy + {y^2}}}{{{x^2} - xy + {y^2}}} = ? \cr
& \Rightarrow {\text{ }}\frac{{{x^2} + {y^2} + 2xy - xy}}{{{x^2} + {y^2} - 2xy + xy}} \cr
& \Rightarrow {\text{ }}\frac{{{{\left( {x + y} \right)}^2} - xy}}{{{{\left( {x - y} \right)}^2} + xy}} = ? \cr
& {\text{Now,}} \cr
& {\text{ }}x + y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} + \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr
& \Rightarrow x + y = \frac{{{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2} + {{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}}}{{{{\sqrt 5 }^2} - {{\sqrt 3 }^2}}} \cr
& \Rightarrow {\text{ }}x + y = \frac{{2\left( {{{\sqrt 5 }^2} + {{\sqrt 3 }^2}} \right)}}{{5 - 3}} \cr
& \Rightarrow x + y = 8\,.......(i) \cr
& Again, \cr
& x - y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} - \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr
& \Rightarrow {\text{ }}x - y = \frac{{4 \times \sqrt 5 \times \sqrt 3 }}{2} \cr
& \Rightarrow x - y = 2\sqrt {15} ..............(ii) \cr
& {\text{And, }}xy = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr
& \Rightarrow {\text{ }}xy = 1 \cr
& {\text{Substitutes values in the question}}{\text{.}} \cr
& \Rightarrow \frac{{{{\left( {x + y} \right)}^2} - xy}}{{{{\left( {x - y} \right)}^2} + xy}} \cr
& \Rightarrow \frac{{{8^2} - 1}}{{{{\left( {2\sqrt {15} } \right)}^2} + 1}} \cr
& \Rightarrow \frac{{63}}{{61}} \cr} $$

Answer: Option B. -> 2
$$\eqalign{
& {\text{Given,}} \cr
& {x^2} + {y^2} + {z^2} = 2\left( {x + z - 1} \right) \cr
& {\text{Find, }}{x^3} + {y^3} + {z^3} = ? \cr
& \Rightarrow {x^2} + {y^2} + {z^2} = 2\left( {x + z - 1} \right) \cr
& \Rightarrow {x^2} + {y^2} + {z^2} = 2x + 2z - 2 \cr
& \Rightarrow {x^2} + {y^2} + {z^2} = 2x + 2z - 1 - 1 \cr
& \Rightarrow \left( {{x^2} + 1 - 2x} \right) + {y^2} + \left( {{z^2} + 1 - 2z} \right) = 0 \cr
& \Rightarrow {\left( {x - 1} \right)^2} + {y^2} + {\left( {z - 1} \right)^2} = 0 \cr
& \Rightarrow {\left( {x - 1} \right)^2} = 0 \cr
& \Rightarrow x = 1 \cr
& \Rightarrow {y^2} = 0 \cr
& \Rightarrow y = 0 \cr
& \Rightarrow {\left( {z - 1} \right)^2} = 0 \cr
& \Rightarrow z = 1 \cr
& {\text{Value substituted in question,}} \cr
& \Rightarrow {x^3} + {y^3} + {z^3} \cr
& \Rightarrow {1^3} + 0 + {1^3} \cr
& \Rightarrow 2 \cr} $$

Answer: Option B. -> 2
$$\eqalign{
& {\text{Given, }}x + \frac{1}{x} = 1 \cr
& {\text{Find }}\frac{2}{{{x^2} - x + 2}} = ? \cr
& x + \frac{1}{x} = 1 \cr
& {x^2} + 1 = x \cr
& \left( {{x^2} - x} \right) = - 1 \cr
& {\text{Putting value in,}} \cr
& = \frac{2}{{{x^2} - x + 2}} \cr
& = \frac{2}{{ - 1 + 2}} \cr
& = 2 \cr} $$

Answer: Option C. -> $$\frac{1}{{135}}$$
5x + 9y = 5 . . . . . (i)
125x3 + 729y3 = 120 . . . . . (ii)
From equation (i) cubing both sides
⇒ (5x + 9y)3 = 53
⇒ 125x3 + 729y3 + 3 × 5x × 9y(5x + 9y) = 125
⇒ 125x3 + 729y3 + 135xy × 5 = 125
⇒ 120 + 135xy × 5 = 125
⇒ 135xy × 5 = 5
⇒ xy = $$\frac{1}{{135}}$$
∴ product of x & y = $$\frac{1}{{135}}$$

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

Answer: Option D. -> 14
$$\eqalign{
& x - \frac{1}{x} = 2{\text{ to find }}{x^3} - \frac{1}{{{x^3}}} \cr
& \Rightarrow x - \frac{1}{x} = 2 \cr
& \left[ {{\text{Cubing both sides}}} \right] \cr
& \Rightarrow {\left( {x - \frac{1}{x}} \right)^3} = {\left( 2 \right)^3} \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times x \times \frac{1}{x}\left( {x - \frac{1}{x}} \right) = 8 \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times \left( 2 \right) = 8 \cr
& \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 14 \cr} $$

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

Answer: Option C. -> $$\frac{{3xyz}}{{abc}}$$
$$\eqalign{
& x = a\left( {b - c} \right) \cr
& y = b\left( {c - a} \right) \cr
& z = c\left( {a - b} \right) \cr
& {\text{Let, }} \cr
& \frac{x}{a} = b - c{\text{ }}\,\,\,\,\,\,\,\,\frac{x}{a} = {\text{A}} \cr
& \frac{y}{b} = c - a{\text{ }}\,\,\,\,\,\,\,\,\frac{y}{b} = {\text{B}} \cr
& \frac{z}{c} = a - b{\text{ }}\,\,\,\,\,\,\,\,\frac{z}{c} = {\text{C}} \cr
& \therefore {\text{A}} + {\text{B}} + {\text{C}} \cr
& = b - c + c - a + a - b \cr
& = 0 \cr
& \therefore {{\text{A}}^3} + {{\text{B}}^3} + {{\text{C}}^3} = {\text{3ABC}} \cr
& \therefore {\left( {\frac{x}{a}} \right)^3}{\text{ + }}{\left( {\frac{y}{b}} \right)^3} + {\left( {\frac{z}{c}} \right)^3} \cr
& = 3 \times \frac{x}{a} \times \frac{y}{b} \times \frac{z}{c} \cr
& = \frac{{3xyz}}{{abc}} \cr} $$

Answer: Option D. -> 1 : 1 : 1
$$\eqalign{
& {\text{Given, }} \cr
& {a^2} + {b^2} + {c^2} - ab - bc - ca = 0 \cr
& {\text{Find, }}a:b:c = ? \cr
& {\text{According to the question, }} \cr
& {a^2} + {b^2} + {c^2} - ab - bc - ca = 0 \cr
& \Rightarrow {a^2} + {b^2} + {c^2} = ab + bc + ca \cr
& {\text{Let }}a = b = c = 1 \cr
& \Rightarrow {1^2} + {1^2} + {1^2} = 1 \times 1 + \left( {1 \times 1} \right) + \left( {1 \times 1} \right) \cr
& \Rightarrow 3 = 3 \cr
& \Rightarrow {\text{So, ratio of }}a:b:c = 1:1:1 \cr} $$