$$\eqalign{ & \left( {2 + \sqrt 3 } \right)a = \left( {2 - \sqrt 3 } \right)b = 1 \cr & \Rightarrow \frac{1}{a} = \left( {2 + \sqrt 3 } \right) \cr & {\text{By rationals}} \cr & \Rightarrow \frac{1}{b} = \left( {2 - \sqrt 3 } \right) \cr & \Rightarrow \frac{1}{a} + \frac{1}{b} = 2 + \sqrt 3 + 2 - \sqrt 3 \cr & \Rightarrow \frac{1}{a} + \frac{1}{b} = 4 \cr} $$

$$a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}$$
To save your time assume values of a, b, c according to the question.
$$\eqalign{ & {\text{Let }}a = 2,{\text{ }}b = - 1\& c = \frac{1}{2} \cr & 2 + \frac{1}{{ - 1}} = - 1 + \frac{1}{{\frac{1}{2}}} = \frac{1}{2} + \frac{1}{2} \cr & \therefore abc = 2 \times - 1 \times \frac{1}{2} = - 1 \cr} $$

$$\eqalign{ & p + \frac{1}{4} + \sqrt p + {k^2} \cr & = p + \sqrt p + \left( {{k^2} + \frac{1}{4}} \right) \cr & = {\left( {\sqrt p } \right)^2} + 2 \times \frac{1}{2} \times \sqrt p + \left( {{k^2} + \frac{1}{4}} \right) \cr & = {{\text{A}}^2} + {\text{2}} \times {\text{A}} \times {\text{B}} + {{\text{B}}^2} \cr & {\text{A}} = \sqrt p \cr & {{\text{B}}^2} = \left( {{k^2} + \frac{1}{4}} \right) \cr & {\text{B}} = \frac{1}{2} \cr & \therefore {k^2} + \frac{1}{4} = {\left( {\frac{1}{2}} \right)^2} \cr & \Rightarrow {k^2} + \frac{1}{4} = \frac{1}{4} \cr & \Rightarrow {k^2} = 0 \cr & \Rightarrow k = 0 \cr} $$

$$\eqalign{ & \frac{{b - c}}{a}{\text{ + }}\frac{{a + c}}{b}{\text{ + }}\frac{{a - b}}{c} = 1 \cr & a - b + c \ne 0 \cr & {\text{Let }}b = c \cr & \therefore \frac{{b - b}}{a}{\text{ + }}\frac{{a + b}}{b}{\text{ + }}\frac{{a - b}}{b} = 1 \cr & \Rightarrow 0 + \frac{a}{b} + 1 + \frac{a}{b} - 1 = 1 \cr & \Rightarrow \frac{a}{b} + \frac{a}{b} = 1 \cr & \Rightarrow \frac{1}{b} + \frac{1}{b} = \frac{1}{a} \cr & {\text{We take }}b = c \cr & \therefore \boxed{\frac{1}{b} + \frac{1}{c} = \frac{1}{a}} \cr} $$

$$\eqalign{ & {\text{Reciprocal of }}\left( {x + \frac{1}{x}} \right){\text{ }} \cr & = \frac{1}{{\left( {x + \frac{1}{x}} \right)}} \cr & = \frac{x}{{{x^2} + 1}} \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 2 \cr & {\text{Put x = 1}} \cr & \therefore {\text{1 + }}\frac{1}{{\left( 1 \right)}} = 2 \cr & \Rightarrow 2 = 2{\text{ }}\left( {{\text{Satisty}}} \right) \cr & \therefore \left( {{x^2} + \frac{1}{{{x^2}}}} \right)\left( {{x^3} + \frac{1}{{{x^3}}}} \right) \cr & = \left( {1 + 1} \right)\left( {1 + 1} \right) \cr & = 2 \times 2 \cr & = 4 \cr} $$

$$\eqalign{ & {\text{2}}{x^2} - 7x + 12 = 0 \cr & {\text{roots are }}\alpha {\text{ , }}\beta \cr & \therefore \alpha \beta = + \frac{c}{a},\alpha + \beta = \frac{{ - b}}{a} \cr & \therefore \alpha + \beta = + \frac{7}{2},\alpha \beta = \frac{{12}}{2} = 6 \cr & \therefore \frac{\alpha }{\beta } + \frac{\beta }{\alpha } \cr & = \frac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }} \cr & = \frac{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta }}{{\alpha \beta }} \cr & = \frac{{{{\left( {\frac{7}{2}} \right)}^2} - 2 \times 6}}{6} \cr & = \frac{{\frac{{49}}{4} - 12}}{6} \cr & = \frac{{49 - 48}}{{6 \times 4}} \cr & = \frac{1}{{24}} \cr} $$

$$\eqalign{ & \frac{1}{{\root 3 \of 4 + \root 3 \of 2 + 1}} = a\root 3 \of 4 + b\root 3 \of 2 + c \cr & \Rightarrow \frac{1}{{\root 3 \of 4 + \root 3 \of 2 + 1}} = \frac{1}{{{{\left( {{2^{\frac{1}{3}}}} \right)}^2} + {2^{\frac{1}{3}}} + {{\left( 1 \right)}^2}}} \cr & \Rightarrow \therefore {{\text{A}}^3} - {{\text{B}}^3}{\text{ = }}\left( {{\text{A}} - {\text{B}}} \right)\left( {{{\text{A}}^2} + {\text{AB}} + {{\text{B}}^2}} \right) \cr & {\text{Put, A}} = {2^{\frac{1}{3}}}{\text{, B}} = 1 \cr & \Rightarrow \frac{{\left( {{2^{\frac{1}{3}}} - 1} \right)}}{{\left( {{2^{\frac{1}{3}}} - 1} \right)\left( {{{\left( {{2^{\frac{1}{3}}}} \right)}^2} + {2^{\frac{1}{3}}} + {{\left( 1 \right)}^2}} \right)}} \cr & \Rightarrow \frac{{\left( {{2^{\frac{1}{3}}} - 1} \right)}}{{{{\left( {{2^{\frac{1}{3}}}} \right)}^3} - {{\left( 1 \right)}^3}}} \cr & \Rightarrow \left( {{2^{\frac{1}{3}}} - 1} \right) \cr & \therefore {2^{\frac{1}{3}}} - 1 \cr & = a\left( {{2^{\frac{2}{3}}}} \right) + b{\left( 2 \right)^{\frac{1}{3}}} + c \cr & \left( {{\text{Comparing the terms}}} \right) \cr & a = 0 \cr & b = 1 \cr & c = - 1 \cr & \therefore a + b + c \cr & = 0 + 1 - 1 \cr & = 0 \cr} $$

$$\eqalign{ & {\text{ }}x = \root 3 \of {2 + \sqrt 3 } \cr & {x^3} = 2 + \sqrt 3 \cr & \frac{1}{{{x^3}}} = \frac{1}{{2 + \sqrt 3 }} \times \frac{{2 - \sqrt 3 }}{{2 - \sqrt 3 }} = 2 - \sqrt 3 \cr & \therefore {x^3}{\text{ + }}\frac{1}{{{x^3}}} \cr & = 2 + \sqrt 3 + 2 - \sqrt 3 \cr & = 4 \cr} $$

$$\frac{{{p^2} - p}}{{2{p^3} + {p^2}}} + \frac{{{p^2} - 1}}{{{p^2} + 3p}} + \frac{{{p^2}}}{{p + 1}}$$
In such type of question assume values of p
$$\eqalign{ & \therefore {\text{Let }}p = 1 \cr & \therefore \frac{{1 - 1}}{{2 + 1}} + \frac{{1 - 1}}{{1 + 3}} + \frac{1}{{1 + 1}} \cr & = 0 + 0 + \frac{1}{2} \cr & = \frac{1}{2} \cr & {\text{Now check option 'B'}} \cr & \frac{1}{{2{p^2}}} = \frac{1}{2} \cr & {\text{Hence the answer is option 'B'}} \cr} $$

$$\eqalign{ & \frac{x}{y} = \frac{4}{5} \cr & \frac{4}{7} + \frac{{2y - x}}{{2y + x}} \cr & = \frac{4}{7} + \frac{{y\left( {2 - \frac{x}{y}} \right)}}{{y\left( {2 + \frac{x}{y}} \right)}} \cr & = \frac{4}{7} + \frac{{\left( {2 - \frac{4}{5}} \right)}}{{\left( {2 + \frac{4}{5}} \right)}} \cr & = \frac{4}{7} + \frac{{10 - 4}}{{10 + 4}} \cr & = \frac{4}{7} + \frac{6}{{14}} \cr & {\text{ = }}\frac{{8 + 6}}{{14}} \cr & {\text{ = }}\frac{{14}}{{14}} \cr & {\text{ = 1}} \cr} $$

$$\eqalign{ & a + b = 12\,.....{\text{(i)}} \cr & ab = 22\,.....{\text{(ii)}} \cr} $$
Squaring both sides of equation (i)
$$\eqalign{ & \Rightarrow {a^2} + {b^2} + 2ab = 144 \cr & \Rightarrow {a^2} + {b^2} + 2 \times 22 = 144 \cr & \Rightarrow {a^2} + {b^2} = 144 - 44 \cr & \Rightarrow {a^2} + {b^2} = 100 \cr} $$

$$\eqalign{ & {\text{ }}2x + \frac{2}{x} = 3 \cr & \Rightarrow {\text{ }}x + \frac{1}{x} = \frac{3}{2} \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = {\left( {\frac{3}{2}} \right)^3}{\text{ }} \cr & x + \frac{1}{x} = a \cr & {x^3} + \frac{1}{{{x^3}}} = {a^3} - 3a \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\left( {x + \frac{1}{x}} \right) = \frac{{27}}{8} \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3 \times \frac{3}{2} = \frac{{27}}{8} \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = \frac{{27}}{8} - \frac{9}{2} \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = \frac{{ - 9}}{8} \cr & \therefore {x^3} + \frac{1}{{{x^3}}} + 2 \cr & = \frac{{ - 9}}{8} + 2 \cr & = \frac{{ - 9 + 16}}{8} \cr & = \frac{7}{8} \cr} $$

$$\eqalign{ & a + b + c = 15{\text{ }} \cr & {a^2} + {b^2} + {c^2} = 83{\text{ }}\left( {{\text{Given}}} \right) \cr & \therefore a + b + c = 15 \cr & \left( {{\text{Squaring both sides}}} \right){\text{ }} \cr & \Rightarrow {\left( {a + b + c} \right)^2} = {\left( {15} \right)^2}{\text{ }} \cr & \Rightarrow {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = 225 \cr & \Rightarrow 83 + 2\left( {ab + bc + ca} \right) = 225 \cr & \Rightarrow 2\left( {ab + bc + ca} \right) = 142 \cr & \Rightarrow ab + bc + ca = 71 \cr} $$
  $$\therefore {a^3} + {b^3} + {c^3} - 3abc = $$     $$\left( {a + b + c} \right)$$  $$\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$$
$$\eqalign{ & \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 15\left( {83 - 71} \right) \cr & \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 15 \times 12 \cr & \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 180 \cr} $$

$$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 83 \cr & {\text{Subtracting 2 from both sides}} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 83 - 2 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2.x.\frac{1}{x} = 83 - 2 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = 81 \cr & \Rightarrow x - \frac{1}{x} = 9 \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3{\text{ }}\left( {x - \frac{1}{x}} \right) = 729 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3 \times {\text{9}} = 729 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 729 + 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 756 \cr} $$

$$\eqalign{ & {\text{For minimum value of}} \cr & \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \cr & a = b = c \cr & a + b + c = 1{\text{ }}\left( {{\text{Given}}} \right) \cr & \therefore a = b = c = \frac{1}{3} \cr & \frac{1}{a} = \frac{1}{b} = \frac{1}{c} = 3 \cr & \therefore {\text{Minimum value of,}} \cr & = \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \cr & = 3 + 3 + 3 \cr & = 9 \cr} $$

$$\eqalign{ & {x^3} + \frac{3}{x} = 4\left( {{a^3} + {b^3}} \right)\, . . . . . {\text{(i)}} \cr & 3x + \frac{1}{{{x^3}}} = 4\left( {{a^3} - {b^3}} \right)\, . . . . . (ii) \cr & {\text{Equation (i)}} + {\text{(ii)}} \cr & {\left( {x + \frac{1}{x}} \right)^3} = 8{a^3} \cr & \Leftrightarrow x + \frac{1}{x} = 2a\, . . . . . (iii) \cr & Equation{\text{ }}(i) - (ii) \cr & x - \frac{1}{x} = 2b\, . . . . . (iv) \cr & Equation{\text{ }}(iii) - (iv) \cr & 2\left( {a - b} \right) = \frac{2}{x} \cr & \Rightarrow a - b = \frac{1}{x} \cr & \Rightarrow a + b = x \cr & \because {a^2} - {b^2} = \left( {a + b} \right) \times \left( {a - b} \right) \cr & \Rightarrow {a^2} - {b^2} = x \times \frac{1}{x} \cr & \Rightarrow {a^2} - {b^2} = 1 \cr} $$

$$\eqalign{ & {\text{121}}{a^2} + 64{b^2} \cr & = {\left( {11a} \right)^2} + {\left( {8b} \right)^2} + 2 \times 11a \times 8b \cr & = {\left( {11a + 8b} \right)^2} \cr} $$
∴ So, term added to make perfect square = 176 ab

$$\eqalign{ & x + \frac{1}{{x + 1}} = 1 \cr & {\text{Adding both 1 sides}} \cr & \Rightarrow x + 1 + \frac{1}{{x + 1}} = 1 + 1 \cr & \Rightarrow \left( {x + 1} \right) + \frac{1}{{\left( {x + 1} \right)}} = 2 \cr & {\text{Put }}x + 1 = 1 \cr & {\text{And }}\frac{1}{{x + 1}} = 1 \cr & \therefore {\left( {x + 1} \right)^5}{\text{ + }}\frac{1}{{{{\left( {x + 1} \right)}^5}}} \cr & = 1 + 1 \cr & = 2 \cr} $$

$$\eqalign{ & {\text{If }}a + b + c = 0 \cr & {\text{then,}} \cr & {a^3} + {b^3} + {c^3} - 3abc = 0 \cr & \Leftrightarrow {a^3} + {b^3} + {c^3} = 3abc \cr} $$

x = y = 333,     z = 334
⇒ x³ + y³ + z³ - 3xyz = $$\frac{1}{2}$$ (x + y + z) [(x - y)² + (y - z)² + (z - x)²]
⇒ x³ + y³ + z³ - 3xyz = $$\frac{1}{2}$$ (333 + 333 + 334) (333 - 333)² + (333 - 334)² + (334 - 333)²
⇒ x³ + y³ + z³ - 3xyz = $$\frac{1}{2}$$ (1000) (0 + 1 + 1)
⇒ x³ + y³ + z³ - 3xyz = 1000

$$\eqalign{ & {\left( {a + \frac{1}{a}} \right)^2} = 3 \cr & a + \frac{1}{a} = \sqrt 3 \cr & {\text{Take cube on both sides}} \cr & {\left( {a + \frac{1}{a}} \right)^3} = {\left( {\sqrt 3 } \right)^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 0 \cr} $$

$$\eqalign{ & x + y + z = 6 \cr & {x^2} + {y^2} + {z^2} = 20 \cr & \Rightarrow {\left( {x + y + z} \right)^2} = {\left( 6 \right)^2} \cr & \Rightarrow {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + zx} \right) = 36 \cr & \Rightarrow 20 + 2\left( {xy + yz + zx} \right) = 36 \cr & \Rightarrow 2\left( {xy + yz + zx} \right) = 16 \cr & \Rightarrow xy + yz + zx = 8 \cr & \therefore {\text{ }}{x^3} + {y^3} + {z^3} - 3xyz \cr & = \left( {x + y + z} \right)\left( {{\text{ }}{x^2} + {y^2} + {z^2} - xy - zx - yz} \right) \cr & = {x^3} + {y^3} + {z^3} - 3xyz = 6\left( {20 - 8} \right) \cr & = 6\left( {20 - 8} \right) \cr & = 6 \times 12 \cr & = 72 \cr} $$

$$\eqalign{ & x = a - b \cr & y = b - c \cr & z = c - a \cr & \therefore x + y + z \cr & = a - b + b - c + c - a \cr & = 0 \cr & \therefore {x^3} + {y^3} + {z^3} - 3xyz \cr & = \left( {x + y + z} \right)\left( {{\text{ }}{x^2} + {y^2} + {z^2} - xy - zx - yz} \right) \cr & = \left( 0 \right)\left( {{\text{ }}{x^2} + {y^2} + {z^2} - xy - zx - yz} \right) \cr & = 0 \cr} $$

$$\eqalign{ & {\text{ }}x = 3 + 2\sqrt 2 \cr & \Rightarrow {x^2} = {\left( {3 + 2\sqrt 2 } \right)^2} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {x^2} = 9 + 8 + 12\sqrt 2 \cr & \Rightarrow {x^2} = 17 + 12\sqrt 2 \cr & \Rightarrow \frac{1}{{{x^2}}} = \frac{1}{{17 + 12\sqrt 2 }} \times \frac{{17 - 12\sqrt 2 }}{{17 - 12\sqrt 2 }} \cr & \Rightarrow \frac{1}{{{x^2}}} = 17 - 12\sqrt 2 \cr & \therefore {\text{ }}{x^2}{\text{ + }}\frac{1}{{{x^2}}} \cr & = 17 + 12\sqrt 2 + 17 - 12\sqrt 2 \cr & = 34 \cr} $$

$$\eqalign{ & x\left( {3 - \frac{2}{x}} \right) = \frac{3}{x} \cr & \Rightarrow 3x - 2 = \frac{3}{x} \cr & \Rightarrow 3x - \frac{3}{x} = 2 \cr & \Rightarrow 3\left( {x - \frac{1}{x}} \right) = 2 \cr & \Rightarrow x - \frac{1}{x} = \frac{2}{3} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {x^2}{\text{ + }}\frac{1}{{{x^2}}} - 2 = \frac{4}{9} \cr & \Rightarrow {x^2}{\text{ + }}\frac{1}{{{x^2}}} = \frac{4}{9} + 2 \cr & \Rightarrow {x^2}{\text{ + }}\frac{1}{{{x^2}}} = 2\frac{4}{9} \cr} $$

$$\left( {1 - \frac{{2xy}}{{{x^2} + {y^2}}}} \right) \div \left( {\frac{{{x^3} - {y^3}}}{{x - y}} - 3xy} \right)$$
$$ = \left( {\frac{{{x^2} + {y^2} - 2xy}}{{{x^2} + {y^2}}}} \right) \div $$    $$\left( {\frac{{{x^3} - {y^3} - 3xy\left( {x - y} \right)}}{{x - y}}} \right)$$
$$\eqalign{ & = \frac{{{{\left( {x - y} \right)}^2}}}{{{x^2} + {y^2}}} \div \frac{{{{\left( {x - y} \right)}^3}}}{{x - y}} \cr & = \frac{{{{\left( {x - y} \right)}^2}}}{{{x^2} + {y^2}}} \div {\left( {x - y} \right)^2} \cr & = \frac{{{{\left( {x - y} \right)}^2}}}{{{x^2} + {y^2}}} \times \frac{1}{{{{\left( {x - y} \right)}^2}}} \cr & = \frac{1}{{{x^2} + {y^2}}} \cr} $$

$$\eqalign{ & {x^2} + {y^2} + 2x + 1 = 0 \cr & \Rightarrow {\left( {x + 1} \right)^2} + {y^2} = 0 \cr & {\text{Let }}{\left( {x + 1} \right)^2} = 0 \cr & {y^2} = 0 \cr & x = - 1,y = 0 \cr & {\text{Now, }}{x^{31}} + {y^{35}} \cr & = {\left( { - 1} \right)^{31}} + {\left( 0 \right)^{35}} \cr & = - 1 \cr} $$

$$\eqalign{ & a + b = 1 \cr & c + d = 1 \cr & a - b = \frac{d}{c} \cr & or\,\,\frac{1}{{a - b}} = \frac{c}{d} \cr & \Rightarrow \frac{{a + b}}{{a - b}} = \frac{c}{d}\left( {\therefore a + b = 1} \right) \cr & {\text{By C & D rule }} \cr & \Rightarrow \frac{{\left( {a + b} \right) + \left( {a - b} \right)}}{{\left( {a + b} \right) - \left( {a - b} \right)}} = \frac{{c + d}}{{c - d}} \cr & \Rightarrow \frac{{2a}}{{2b}} = \frac{{c + d}}{{c - d}} \cr & \Rightarrow \frac{a}{b} = \frac{{c + d}}{{c - d}} \cr} $$
Now multiply & divide by (c + d)
$$\eqalign{ & \frac{a}{b} = \frac{{\left( {c + d} \right)}}{{\left( {c - d} \right)}} \times \frac{{\left( {c + d} \right)}}{{\left( {c + d} \right)}} = \frac{{{{\left( {c + d} \right)}^2}}}{{{c^2} - {d^2}}} \cr & \Rightarrow \frac{a}{b} = \frac{{{{\left( {c + d} \right)}^2}}}{{\left( {{c^2} - {d^2}} \right)}} \cr & \Rightarrow c + d = 1 \cr & \Rightarrow \frac{a}{b} = \frac{1}{{{c^2} - {d^2}}} \cr & \Rightarrow {c^2} - {d^2} = \frac{b}{a} \cr} $$

$$\eqalign{ & x = 3t\,......(i) \cr & y = \frac{1}{2}\left( {t + 1} \right) \cr & x = 2y \cr & \Rightarrow x = 2 \times \frac{1}{2}\left( {t + 1} \right) \cr & \Rightarrow x = t + 1\,......(ii) \cr & \therefore 3t = t + 1 \cr & \left( {{\text{From equation (i) and (ii)}}} \right) \cr & \Rightarrow 2t = 1 \cr & \Rightarrow t = \frac{1}{2} \cr} $$