$$\eqalign{ & a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a} \cr & {\text{Put , }}a = \frac{1}{2},b = 2,c = - 1 \cr & \Rightarrow \frac{1}{2} + \frac{1}{2} = 2 - 1 = - 1 + 2 \cr & \Rightarrow 1 = 1 = 1 \cr & abc = \frac{1}{2} \times 2 \times - 1 \cr & \boxed{abc = - 1} \cr & {\text{Again put}} \cr & a = - \frac{1}{2},b = - 2,c = 1 \cr & \Rightarrow a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a} \cr & \Rightarrow - \frac{1}{2} - \frac{1}{2} = - 2 + 1 = 1 - 2 \cr & \Rightarrow - 1 = - 1 = - 1 \cr & {\text{Equation satisfied}} \cr & \Rightarrow abc = - \frac{1}{2} \times - 2 \times 1 \cr & \boxed{abc = + 1} \cr & {\text{So, }}abc{\text{ can be }} - 1{\text{ and }} + 1 \cr} $$

$$\eqalign{ & {a^3} - 7a - 6 \cr & \left( {a + 1} \right)\left( {{a^2} - a - 6} \right) \cr & \left( {a + 1} \right)\left( {a + 2} \right)\left( {a - 3} \right) \cr & {\text{Now}},{\text{Sum of factors}} \cr & \left( {a + 1} \right) + a + 2 + a - 3 = 3a \cr} $$

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

$$\eqalign{ & a + b = 10{\text{ and }}ab = 21 \cr & \left( {a + b} \right) = 10 \cr & \Rightarrow {a^2} + {b^2} + 2ab = 100 \cr & \Rightarrow {a^2} + {b^2} = 100 - 2ab \cr & \Rightarrow {a^2} + {b^2} = 100 - 2 \times 21 \cr & \Rightarrow {a^2} + {b^2} = 100 - 42 \cr & {a^2} + {b^2} = 58\,.........(i) \cr & {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab \cr & {\left( {a - b} \right)^2} = 58 - 2 \times 21 \cr & \left[ {{\text{from equation (i)}}} \right] \cr & = {\text{58}} - {\text{42}} \cr & {\left( {a - b} \right)^2} = 16 \cr} $$

$$\eqalign{ & 2x + \frac{1}{{4x}} = 1 \cr & {\text{Dividing by 2 both side}} \cr & \Rightarrow x + \frac{1}{{8x}} = \frac{1}{2} \cr & {\text{Squaring both side }} \cr & \Rightarrow {x^2} + \frac{1}{{64{x^2}}} + 2 \times x \times \frac{1}{{8x}} = \frac{1}{4} \cr & \Rightarrow {x^2} + \frac{1}{{64{x^2}}} + \frac{1}{4} = \frac{1}{4} \cr & \Rightarrow {x^2} + \frac{1}{{64{x^2}}} = \frac{1}{4} - \frac{1}{4} \cr & \Rightarrow {x^2} + \frac{1}{{64{x^2}}} = 0 \cr} $$

$$\eqalign{ & \sqrt x - \sqrt y = 1\,......(i) \cr & \sqrt x + \sqrt y = 17\,.....(ii) \cr & {\text{From equation (i) and (ii)}} \cr & \sqrt x = 9,\sqrt y = 8 \cr & {\text{So, }}\sqrt {xy} = 9 \times 8 = 72 \cr} $$

$$\eqalign{ & {\text{ }}x = \sqrt a + \frac{1}{{\sqrt a }} \cr & {\text{ }}y = \sqrt a - \frac{1}{{\sqrt a }} \cr & {\text{Put }}a = 4 \cr & x = 2 + \frac{1}{2} = \frac{5}{2} \cr & y = 2 - \frac{1}{2} = \frac{3}{2} \cr & {\text{Then, }}{x^4} + {y^4} - 2{x^2}{y^2}{\text{ }} \cr & = {\left( {{x^2} - {y^2}} \right)^2} \cr & = {\left( {\frac{{25}}{4} - \frac{9}{4}} \right)^2} \cr & = {\text{ 16}} \cr} $$

$$\eqalign{ & x = \root 3 \of {{x^2} + 11} - 2 \cr & \Rightarrow x + 2 = \root 3 \of {{x^2} + 11} \cr & \Rightarrow {\text{Taking cube on both side}} \cr & \Rightarrow {\left( {x + 2} \right)^3} = {x^2} + 11 \cr & \Rightarrow {x^3} + 8 + 6x\left( {x + 2} \right) = {x^2} + 11 \cr & \Rightarrow {x^3} + 8 + 6{x^2} + 12x = {x^2} + 11 \cr & \Rightarrow {x^3} + 5{x^2} + 12x = 3 \cr} $$

$$\eqalign{ & 2x - \frac{1}{{2x}} = 5 \cr & {\text{Divide by 2 both side}} \cr & x - \frac{1}{{4x}} = \frac{5}{2} \cr & {\text{Squaring both side}} \cr & \Rightarrow {x^2} + \frac{1}{{16{x^2}}} - 2 \times x \times \frac{1}{{4x}} = \frac{{25}}{4} \cr & \Rightarrow {x^2} + \frac{1}{{16{x^2}}} - \frac{1}{2} = \frac{{25}}{4} \cr & \Rightarrow {x^2} + \frac{1}{{16{x^2}}} = \frac{{25}}{4} + \frac{1}{2} \cr & \Rightarrow {x^2} + \frac{1}{{16{x^2}}} = \frac{{27}}{4} \cr & {\text{So, }} \cr & {x^2} + \frac{1}{{16{x^2}}} - 2 \cr & = \frac{{27}}{4} - 2 \cr & = \frac{{19}}{4} \cr} $$

$$\eqalign{ & {\text{Given that,}} \cr & {\text{ }}x = \frac{1}{{\left( {\sqrt 2 + 1} \right)}}{\text{ }} \cr & {\text{Then, }}{x^2} + 2x - 1 \cr & = {x^2} + 2x - 1 + 1 - 1 \cr & = {x^2} + 2x + 1 - 2 \cr & = {\left( {x + 1} \right)^2} - 2 \cr & {\text{Now put the value of }}x \cr & = {\left( {\frac{1}{{\left( {\sqrt 2 + 1} \right)}} + 1} \right)^2} - 2 \cr & = {\left( {\frac{{1 + \sqrt 2 + 1}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr & = {\left( {\frac{{\sqrt 2 + 2}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr & = {\left( {\frac{{\left( {\sqrt 2 + 1} \right) \times \sqrt2}}{{\sqrt 2 + 1}}} \right)^2} - 2 \cr & = {\left( {\sqrt 2 } \right)^2} - 2 \cr & = 2 - 2 \cr & = 0 \cr} $$

$$\eqalign{ & {a^2} + {b^2} = ab\,........(i) \cr & {a^2} + {b^2} - ab = 0 \cr & \because {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)\,.....(ii) \cr & {\text{From equation (i) and (ii)}} \cr & \Rightarrow {a^3} + {b^3} = \left( {a + b} \right)\left( 0 \right) \cr & \Rightarrow {a^3} + {b^3} = 0 \cr} $$

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

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

$$\eqalign{ & \because x = \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr & \Rightarrow x = \frac{{3 + 1}}{{\sqrt 3 }} \cr & \Rightarrow x = \frac{4}{{\sqrt 3 }} \cr & {\text{So, }}\left( {x - \frac{{\sqrt {126} }}{{\sqrt {42} }}} \right)\left( {x - \frac{1}{{x - \frac{{2\sqrt 3 }}{3}}}} \right) \cr & = \left( {\frac{4}{{\sqrt 3 }} - \frac{{\sqrt {126} }}{{\sqrt {42} }}} \right)\left( {\frac{4}{{\sqrt 3 }} - \frac{1}{{\frac{4}{{\sqrt 3 }} - \frac{2}{{\sqrt 3 }}}}} \right) \cr & = \left( {\frac{{4\sqrt {42} - \sqrt {126} \times \sqrt 3 }}{{\sqrt {3 \times } \sqrt {42} }}} \right)\left( {\frac{4}{{\sqrt 3 }} - \frac{1}{{\frac{2}{{\sqrt 3 }}}}} \right) \cr & = \frac{{4\sqrt {42} - 3\sqrt {42} }}{{\sqrt {3 \times } \sqrt {42} }}\left( {\frac{4}{{\sqrt 3 }} - \frac{{\sqrt 3 }}{2}} \right) \cr & = \left( {\frac{{\sqrt {42} }}{{\sqrt {3 \times } \sqrt {42} }}} \right)\left( {\frac{{8 - 3}}{{2\sqrt 3 }}} \right) \cr & = \frac{1}{{\sqrt 3 }} \times \frac{5}{{2\sqrt 3 }} \cr & = \frac{5}{6} \cr} $$

$$\eqalign{ & \frac{{2 + a}}{a} + \frac{{2 + b}}{b} + \frac{{2 + c}}{c} = 4 \cr & \Rightarrow \frac{{2bc + abc + 2ac + abc + 2ab + abc}}{{abc}} = 4 \cr & \Rightarrow \frac{{2\left( {bc + ab + ca} \right)}}{{abc}} + \frac{{3abc}}{{abc}} = 4 \cr & \Rightarrow \frac{{2\left( {bc + ab + ca} \right)}}{{abc}} + 3 = 4 \cr & \Rightarrow \boxed{\frac{{ab + bc + ca}}{{abc}} = \frac{1}{2}} \cr} $$

$$\eqalign{ & {a^3} = 117 + {b^3}{\text{ , }}a = 3 + b \cr & {a^3} - {b^3}{\text{ = 117}}\,......{\text{(i)}} \cr & a - b = 3\,........(ii) \cr & {\text{Put }}a = 5,{\text{ }}b = 2 \cr & {\text{Both equation satisfy}} \cr & {\text{Now, }}a + b = 5 + 2 = 7 \cr} $$

$$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 98 \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} + 2 = 100 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 100 \cr & \Rightarrow x + \frac{1}{x} = 10 \cr & {\text{Cubing}}\,{\text{both}}\,{\text{sides}} \cr & \Rightarrow {\text{ }}{x^3} + \frac{1}{{{x^3}}} = {10^3} - 3 \times 10 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 1000 - 3 \times 10 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 970 \cr} $$

$$\eqalign{ & {p^2} + \frac{1}{{{p^2}}} = 47 \cr & {\text{On adding 2 both side}} \cr & {p^2} + \frac{1}{{{p^2}}} + 2 = 47 + 2 \cr & \Rightarrow {\left( {p + \frac{1}{p}} \right)^2} = 49 \cr & \Rightarrow \left( {p + \frac{1}{p}} \right) = 7 \cr & \Rightarrow p + \frac{1}{p} = 7 \cr} $$

$$\eqalign{ & {\text{Given, }}x + \frac{1}{x} = \sqrt {13} {\text{ }} \cr & {\text{then ,}}\frac{{3x}}{{\left( {{x^2} - 1} \right)}} \cr & = \frac{3}{{x - \frac{1}{x}}}\,.............(i) \cr & {\text{Now, }}x + \frac{1}{x} = \sqrt {13} \cr & {\text{On squaring both side}} \cr & = {x^2} + \frac{1}{{{x^2}}} \cr & = 13 - 2 \cr & = 11 \cr & = {x^2} + \frac{1}{{{x^2}}} - 2 \cr & = 11 - 2 \cr & = 9 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = 9 \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^2} = {3^2} \cr & \Rightarrow x - \frac{1}{x} = 3 \cr & {\text{Put this value in equation (i)}} \cr & \Rightarrow \frac{3}{{x - \frac{1}{x}}} \cr & \Rightarrow \frac{3}{3} \cr & \Rightarrow 1 \cr} $$

$$\eqalign{ & p\left( {{p^2} + 3p + 3} \right) \cr & = {p^3} + 3{p^2} + 3p + 1 - 1 \cr} $$
(On adding and subtracting one both side)
$$\eqalign{ & = {\left( {p + 1} \right)^3} - 1 \cr & = {\left( {99 + 1} \right)^3} - 1 \cr & = {100^3} - 1 \cr & = 1000000 - 1 \cr & = 999999 \cr} $$

$$\eqalign{ & {a^3} - {b^3} = 61 \cr & \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) = 61 \cr & a - b = 1{\text{ }}\left( {{\text{ }}Given} \right) \cr & 1 \times {a^2} + ab + {b^2} = 61\,......(i) \cr & {\text{Now,}}a - b = 1 \cr & {\text{On squaring both sides}} \cr & {a^2} + {b^2} - 2ab = 1 \cr & {a^2} + {b^2} + ab - 3ab = 1 \cr & {\text{From equation (i)}} \cr & \Rightarrow 61 - 3ab = 1 \cr & \Rightarrow 3ab = 60 \cr & \Rightarrow ab = \frac{{60}}{3} \cr & \Rightarrow ab = 20 \cr} $$

$$\eqalign{ & {\text{If }}{x^2} + 5x + 6 = 0 \cr & {\text{then, }}{x^2} + 6 = - 5x \cr & {\text{So,}}\frac{{2x}}{{{x^2} + 6 - 7x}} \cr & = \frac{{2x}}{{ - 5x - 7x}} \cr & = \frac{{2x}}{{ - 12x}} \cr & = - \frac{1}{6}{\text{ }} \cr} $$

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

$$\eqalign{ & {\text{Given , }}x = y + z \cr & \therefore x - y - z = 0 \cr & {\text{Then }}{x^3} - {y^3} - {z^3} = 3xyz \cr} $$

$$\eqalign{ & \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + \frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} \cr & = \frac{{1 - d + 1 - a + 1 - b + 1 - c}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = \frac{{4 - \left( {a + b + c + d} \right)}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = \frac{{4 - 4}}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} \cr & = 0 \cr} $$

$${x^5} - 12{x^4} + 12{x^3} - 12{x^2} + 12x - 1$$
$$ = {x^5} - 11{x^4} - {x^4} + 11{x^3} + {x^3} - 11{x^2} - $$       $$\,{x^2} + $$ $$11x \,+\, $$ $$x - $$ $$1$$
$${\text{Put }}x = 11$$
$$ = {11^5} - {11.11^4} - {11^4} + {11.11^3} + {11^3} - $$       $${11.11^2} -\, $$ $$\,{11^2} \,+\, $$ $$11.11 +\, $$ $$11 - $$ $$1$$
$$\eqalign{ & = 11 - 1 \cr & = 10 \cr} $$

$$\eqalign{ & a + b = 5{\text{ }} \cr & \Leftrightarrow {a^2} + {b^2} + 2ab = 25\,.....(i) \cr & a - b = 3 \cr & \Leftrightarrow {a^2} + {b^2} - 2ab = 9\,.....(ii) \cr & {\text{From equation (i) and (ii)}} \cr & \Leftrightarrow {\text{2}}\left( {{a^2} + {b^2}} \right) = 34 \cr & \Leftrightarrow {a^2} + {b^2} = 17 \cr} $$