$$\eqalign{ & {a^2} + {b^2} = 2{\text{ }} \cr & {c^2} + {d^2} = 1 \cr & {\text{Put values of }}a,{\text{ }}b,{\text{ }}c,{\text{ }}d \cr & {\text{Take ,}}a = b = 1 \cr & c = 1 \cr & d = 0 \cr & \Rightarrow {\left( {ad - bc} \right)^2}{\text{ + }}{\left( {ac - bd} \right)^2} \cr & \Rightarrow {\left( {0 - 1} \right)^2} + {\left( {1 + 0} \right)^2} \cr & \Rightarrow {\left( { - 1} \right)^2} + {\left( 1 \right)^2} \cr & \Rightarrow 2 \cr} $$

$$\eqalign{ & x \propto \frac{1}{{{y^2} - 1}}{\text{ }}\left( {{\text{Given}}} \right) \cr & x = k \times \frac{1}{{{y^2} - 1}}\left( {k{\text{ is constant}}} \right) \cr & {\text{Now }}x = 24{\text{ when }}y = 10{\text{ given}} \cr & \Rightarrow 24 = k \times \frac{1}{{{{\left( {10} \right)}^2} - 1}} \cr & \Rightarrow 24 = \frac{k}{{99}} \cr & \Rightarrow k = 24 \times 99 \cr & x = ? \cr & y = 5 \cr & \Rightarrow x = 24 \times 99 \times \frac{1}{{25 - 1}} \cr & \Rightarrow x = 24 \times 99 \times \frac{1}{{24}} \cr & \Rightarrow x = 99 \cr} $$

$$\eqalign{ & {x^2} + {y^2} + 2x + 1 = 0 \cr & \Rightarrow {x^2} + 2x + 1 + {y^2} = 0 \cr & \Rightarrow {\left( {x + 1} \right)^2} + {y^2} = 0 \cr} $$
Hence both terms are squares and there addition is zero.
So, it can be possible only when both terms are zeros.
$$\eqalign{ & \therefore x + 1 = 0 \cr & \Rightarrow x = - 1 \cr & y = 0 \cr & \therefore {x^{31}} + {y^{35}} \cr & \Rightarrow {\left( { - 1} \right)^{31}} + {\left( 0 \right)^{35}} \cr & \Rightarrow - 1 \cr} $$

$$\eqalign{ & a = 2.361 \cr & b = 3.263 \cr & c = 5.624 \cr & a + b - c = 0 \cr & 2.361 + 3.263 - 5.624 = 0 \cr & 0 = 0 \cr & \therefore {a^3} + {b^3} - {c^3} + 3abc \cr & \Rightarrow 0 \cr} $$

$$\eqalign{ & {\text{ }}x = \frac{{4ab}}{{a + b}}{\text{ }} \cr & \Rightarrow \frac{x}{{2a}} = {\text{ }}\frac{{2b}}{{a + b}}{\text{ }} \cr & \frac{{x + 2a}}{{x - 2a}} \cr & = \frac{{2b + a + b}}{{2b - a - b}} \cr & = \frac{{3b + a}}{{b - a}} \cr} $$
( By Componendo and Dividendo rule )
$$\eqalign{ & \Rightarrow {\text{again}}\frac{x}{{2b}} = \frac{{2a}}{{a + b}} \cr & \frac{{x + 2b}}{{x - 2b}} \cr & = \frac{{2a + a + b}}{{2a - a - b}} \cr & = \frac{{3a + b}}{{a - b}} \cr & \Rightarrow \frac{{x + 2a}}{{x - 2a}}{\text{ + }}\frac{{x + 2b}}{{x - 2b}} \cr & \Rightarrow \frac{{3b + a}}{{b - a}} - \frac{{3a + b}}{{a - b}} \cr & \Rightarrow \frac{{3b + a - 3a - b}}{{b - a}} \cr & \Rightarrow \frac{{2b - 2a}}{{b - a}} \cr & \Rightarrow \frac{{2\left( {b - a} \right)}}{{\left( {b - a} \right)}} \cr & \Rightarrow 2 \cr} $$

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

a² + b² + c² = 2(a - b - c) - 3
⇒ a² + b² + c² = 2a - 2b - 2c - 3
⇒ a² + b² + c² - 2a + 2b +2c + 1 + 1 + 1 = 0
⇒ (a² - 2a + 1) + (b² + 2b + 1) + (c² + 2c + 1) = 0
⇒ (a - 1)² + (b + 1)² + (c + 1)² = 0
a = 1
b = -1
c = -1
∴ 2a - 3b + 4c
= 2 × 1 - 3 × (-1) + 4 × (-1)
= 2 + 3 - 4
= 1

$$\eqalign{ & {\left( {3a + 1} \right)^2} + {\left( {b - 1} \right)^2} + {\left( {2c - 3} \right)^2} = 0 \cr & {\left( {3a + 1} \right)^2} = 0 \cr & \Rightarrow 3a = - 1 \cr & \Rightarrow a = - \frac{1}{3} \cr & {\left( {b - 1} \right)^2} = 0 \cr & \Rightarrow b - 1 = 0 \cr & \Rightarrow b = 1 \cr & {\left( {2c - 3} \right)^2} = 0 \cr & \Rightarrow c = \frac{3}{2} \cr & \therefore 3a + b + 2c \cr & = 3 \times - \frac{1}{3} + 1 + \frac{3}{2} \times 2 \cr & = - 1 + 1 + 3 \cr & = 3 \cr} $$

$$\eqalign{ & \frac{p}{a} + \frac{q}{b} + \frac{r}{c} = 1 \cr & \frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0{\text{ }} \cr & \Rightarrow \frac{p}{a} = x,{\text{ }}\frac{q}{b} = y,{\text{ }}\frac{r}{c} = z \cr & \Rightarrow \left( {x + y + z} \right) = 1 \cr & {\text{Squaring the both sides}} \cr & \Rightarrow {x^2} + {y^2} + {z^2} + 2\left( {xy + yz + zx} \right) = 1 \cr & {\text{and }}\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 0 \cr & \Rightarrow \frac{{xy + yz + zx}}{{xyz}} = 0 \cr & \Rightarrow xy + yz + zx = 0 \cr & \therefore {x^2} + {y^2} + {z^2} = 1 \cr & {\text{So, }}\frac{{{p^2}}}{{{a^2}}} + \frac{{{q^2}}}{{{b^2}}} + \frac{{{r^2}}}{{{c^2}}} = 1 \cr} $$

$$\eqalign{ & {x^2} + \frac{1}{{{x^2}}} = 66 \cr & \text{Subtract 2 from both sides} \cr & \Rightarrow {x^2} + \frac{1}{{{x^2}}} - 2 = 66 - 2 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = 64 \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^2} = {\left( 8 \right)^2} \cr & \Rightarrow x - \frac{1}{x} = \pm 8 \cr & \Rightarrow \frac{{{x^2} - 1 + 2x}}{x} \cr & \Rightarrow \frac{{\frac{{{x^2}}}{x} - \frac{1}{x} + \frac{{2x}}{x}}}{{\frac{x}{x}}} \cr & \Rightarrow \frac{{\left( {x - \frac{1}{x}} \right) + 2}}{1} \cr & {\text{When , }}x - \frac{1}{x} = + 8 \cr & {\text{Then,}} \cr & \left( {x - \frac{1}{x}} \right) + 2 = 8 + 2 = 10 \cr & {\text{When }}x - \frac{1}{x} = - 8 \cr & \Rightarrow x - \frac{1}{x} = - 8 + 2 \cr & \Rightarrow x - \frac{1}{x} = - 6 \cr & \therefore \left( {10, - 6} \right) \cr} $$

$${a^2} + a + 1 = 0$$
\[\left[ \begin{array}{l} {a^3} + {1^3} = \left( {a + 1} \right)\left( {{a^2} + a + 1} \right)\\ {a^3} - {1^3} = \left( {a - 1} \right)\left( {{a^2} + a + 1} \right) \end{array} \right]\]
$$\eqalign{ & \therefore \left( {{a^3} - 1} \right) = \left( {a - 1} \right) \times 0 \cr & \Rightarrow {a^3} - 1 = 0 \cr & \Rightarrow {a^3} = 1 \cr & \Rightarrow {\left( {{a^3}} \right)^3} = {1^3} \cr & \Rightarrow {a^9} = 1 \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & x = - 2k{\text{ and }}y = 1 - 3k \cr & \therefore {\text{For }}x = y \cr & \Rightarrow - 2k = 1 - 3k \cr & \Rightarrow k = 1 \cr} $$

$$\eqalign{ & {\text{ }}x + \frac{1}{x} = 5 \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {x + \frac{1}{x}} \right)^3} = {\left( 5 \right)^3} \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} + 3 \times 5 = 125 \cr & \Rightarrow {x^3}{\text{ + }}\frac{1}{{{x^3}}} = 110 \cr & \therefore {\text{Squaring both sides}} \cr & \Rightarrow {\left( {{x^3}{\text{ + }}\frac{1}{{{x^3}}}} \right)^2} = {\left( {110} \right)^2} \cr & \Rightarrow {x^6}{\text{ + }}\frac{1}{{{x^6}}} + 2 = 12100 \cr & \Rightarrow {x^6}{\text{ + }}\frac{1}{{{x^6}}} = 12100 - 2 \cr & \Rightarrow {x^6}{\text{ + }}\frac{1}{{{x^6}}} = 12098 \cr} $$

$$\eqalign{ & x = \sqrt 3 - \frac{1}{{\sqrt 3 }}{\text{ and }}y = \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr & \Rightarrow \frac{{{x^2}}}{y} + \frac{{{y^2}}}{x} \cr & = \frac{{{x^3} + {y^3}}}{{xy}} \cr & = \frac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)}}{{xy}} \cr & \therefore x + y \cr & = \sqrt 3 - \frac{1}{{\sqrt 3 }} + \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr & = 2\sqrt 3 \cr & \therefore xy \cr & = \sqrt 3 - \frac{1}{{\sqrt 3 }} \times \sqrt 3 + \frac{1}{{\sqrt 3 }} \cr & = 3 - \frac{1}{3} \cr & = \frac{8}{3} \cr & \Rightarrow \frac{{\left( {x + y} \right)\left( {{x^2} + {y^2} + 2xy - 2xy - xy} \right)}}{{xy}} \cr & \Rightarrow \frac{{\left( {x + y} \right)\left( {{{\left( {x + y} \right)}^2} - 3xy} \right)}}{{xy}} \cr & \Rightarrow \frac{{2\sqrt 3 \left( {{{\left( {2\sqrt 3 } \right)}^2} - 3 \times \frac{8}{3}} \right)}}{{\frac{8}{3}}} \cr & \Rightarrow \frac{{2\sqrt 3 \left( {12 - 8} \right)}}{{\frac{8}{3}}} \cr & \Rightarrow \frac{{2 \times 3\sqrt 3 \left( 4 \right)}}{8} \cr & \Rightarrow 3\sqrt 3 \cr} $$

$$\eqalign{ & \frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}} + \frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}} + \frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}} + \frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}} \cr & {\text{Put }} \cr & a = 0 \cr & b = 0 \cr & c = 2 \cr & d = 2 \cr & \therefore a + b + c + d = 4 \cr & \Rightarrow 0 + 0 + 2 + 2 = 4 \cr & \Rightarrow 4 = 4\left( {{\text{ satisfy}}} \right) \cr & \frac{1}{{\left( {1 - 0} \right)\left( {1 - 0} \right)\left( {1 - 2} \right)}} + \frac{1}{{\left( {1 - 0} \right)\left( {1 - 2} \right)\left( {1 - 2} \right)}} + \frac{1}{{\left( {1 - 2} \right)\left( {1 - 2} \right)\left( {1 - 0} \right)}} + \frac{1}{{\left( {1 - 2} \right)\left( {1 - 0} \right)\left( {1 - 0} \right)}} \cr & = \frac{1}{{ - 1}} + \left( {\frac{1}{{ + 1}}} \right) + \frac{1}{{ - 1 \times - 1}} + \frac{1}{{ - 1}} \cr & = - 1 + 1 + 1 - 1 \cr & = 0 \cr} $$

$$\eqalign{ & {a^2} + {b^2} + 2b + 4a + 5 = 0 \cr & \Rightarrow {a^2} + {b^2} + 2b + 4a + 4 + 1 = 0 \cr & \Rightarrow {a^2} + 4a + 4 + {b^2} + 2b + 1 = 0 \cr & \Rightarrow {\left( {a + 2} \right)^2} + {\left( {b + 1} \right)^2} = 0 \cr & a + 2 = 0{\text{ }} \Rightarrow {\text{ a}} = - 2 \cr & b + 1 = 0\,\,\,\, \Rightarrow \,\,\,\,b = - 1 \cr & \frac{{a - b}}{{a + b}} \Rightarrow \frac{{ - 2 + 1}}{{ - 2 - 1}} \cr & \Rightarrow \frac{{ - 1}}{{ - 3}} = \frac{1}{3} \cr} $$

$$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$   $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$    $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$
Now,
$$ \Rightarrow \frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} \times \frac{{\left( {a - b} \right)}}{{a - b}}$$
Multiply divide by (a - b) in 1^st term
Multiply divide by (b - c) in 2^nd term
Multiply divide by (c - a) in 3^rd term
$$ \Rightarrow \frac{{{{\left( {a - b} \right)}^2}\left( {a - b} \right)}}{{\left( {b - c} \right)\left( {c - a} \right)\left( {a - b} \right)}} + $$     $$\frac{{{{\left( {b - c} \right)}^2}\left( {b - c} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} + $$     $$\frac{{{{\left( {c - a} \right)}^2}\left( {c - a} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}$$
$$\eqalign{ & {\text{Let, }}a - b = x \cr & b - c = y \cr & c - a = z \cr & \therefore x + y + z = 0 \cr & \therefore {x^3} + {y^3} + {z^3} = 3xyz \cr & \therefore {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} \cr & = 3\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right) \cr & \therefore \frac{{3\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}}{{\left( {a - b} \right)\left( {b - c} \right)\left( {c - a} \right)}} \cr & = 3 \cr} $$

$$\eqalign{ & \frac{{{{\left( {x + 1} \right)}^3} - {{\left( {x - 1} \right)}^3}}}{{{{\left( {x + 1} \right)}^2} - {{\left( {x - 1} \right)}^2}}} = 2 \cr & {{\text{A}}^3} - {{\text{B}}^3} = \left( {{\text{A}} - {\text{B}}} \right)\left( {{{\text{A}}^2} + {\text{AB}} + {{\text{B}}^2}} \right) \cr & {{\text{A}}^2} - {{\text{B}}^2} = \left( {{\text{A}} - {\text{B}}} \right)\left( {{\text{A}} + {\text{B}}} \right) \cr} $$
$$ \Rightarrow \frac{{\left( {x + 1 - x + 1} \right)\left\{ {{{\left( {x + 1} \right)}^2} + \left( {x - 1} \right)\left( {x + 1} \right) + {{\left( {x - 1} \right)}^2}} \right\}}}{{\left( {x + 1 - x + 1} \right)\left( {x + 1 + x - 1} \right)}} = 2$$

$$\eqalign{ & \Rightarrow \frac{{\left( {{x^2} + 1 + 2x + {x^2} - 1 + {x^2} + 1 - 2x} \right)}}{{\left( {2x} \right)}} = 2 \cr & \Rightarrow \frac{{3{x^2} + 1}}{{2x}} = 2 \cr & \Rightarrow 3{x^2} + 1 = 4x \cr & \Rightarrow 3{x^2} - 4x + 1 = 0 \cr & \Rightarrow 3{x^2} - 3x - x + 1 = 0 \cr & \Rightarrow 3x\left( {x - 1} \right) - 1\left( {x - 1} \right) = 0 \cr & \Rightarrow \left( {3x - 1} \right)\left( {x - 1} \right) = 0 \cr & 3x - 1 = 0 \cr & \Leftrightarrow x = \frac{1}{3} \cr & x - 1 = 0 \cr & \text{for } x = 1 = \frac{1}{1} \cr & {\text{By adding numinator and denominator }} \cr & 1 + 1 = 2 \cr & {\text{No option is satisfied}} \cr & \therefore x = \frac{1}{3} \cr & \Rightarrow x = 1 + 3 \cr & \Rightarrow x = 4 \cr} $$

$$x = \sqrt 5 + 2$$
Rationalization of denominator.
$$\eqalign{ & \Rightarrow \frac{1}{x} = \frac{1}{{\sqrt 5 + 2}} \times \frac{{\sqrt 5 - 2}}{{\sqrt 5 - 2}} \cr & \Rightarrow \frac{1}{x} = \frac{{\sqrt 5 - 2}}{{5 - 4}} \cr & \Rightarrow \frac{1}{x} = \sqrt 5 - 2 \cr & \Rightarrow x - \frac{1}{x} = \sqrt 5 + 2 - \sqrt 5 + 2 \cr & \Rightarrow x - \frac{1}{x} = 4 \cr & \therefore \frac{{2{x^2} - 3x - 2}}{{3{x^2} - 4x - 3}} \cr} $$
Multiply by $$\frac{1}{x}$$ in numerator and denominator
$$\eqalign{ & = \frac{{\frac{{2{x^2}}}{x} - \frac{{3x}}{x} - \frac{2}{x}}}{{\frac{{3{x^2}}}{x} - \frac{{4x}}{x} - \frac{3}{x}}} \cr & = \frac{{2x - \frac{2}{x} - 3}}{{3x - \frac{3}{x} - 4}} \cr & = \frac{{2\left( {x - \frac{1}{x}} \right) - 3}}{{3\left( {x - \frac{1}{x}} \right) - 4}} \cr & = \frac{{2 \times 4 - 3}}{{3 \times 4 - 4}} \cr & = \frac{{8 - 3}}{{12 - 4}} \cr & = \frac{5}{8} \cr & = 0.625 \cr} $$

$$\eqalign{ & {\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}} + {c^{\frac{1}{3}}} = 0 \cr & \Rightarrow {\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}} = - {c^{\frac{1}{3}}} \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {{\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}}} \right)^3} = {\left( { - {c^{\frac{1}{3}}}} \right)^3} \cr & \Rightarrow {\text{ }}a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {{\text{ }}{a^{\frac{1}{3}}} + {b^{\frac{1}{3}}}} \right) = - c \cr & \Rightarrow {\text{ }}a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {{\text{ }} - {c^{\frac{1}{3}}}} \right) = - c \cr & \Rightarrow a + b + c = 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}{c^{\frac{1}{3}}} \cr & {\text{Again taking cube }} \cr & \Rightarrow {\left( {a + b + c} \right)^3} = 27abc \cr} $$

$$\eqalign{ & x - \frac{1}{x} = 1{\text{ }} \cr & \Rightarrow \frac{{{x^4} - \frac{1}{{{x^2}}}}}{{3{x^2} + 5x - 3}} \cr & {\text{Divide and multiply by }}x \cr & \Rightarrow \frac{{\frac{{{x^4}}}{x} - \frac{1}{{{x^3}}}}}{{\frac{{3{x^2}}}{x} + \frac{{5x}}{x} - \frac{3}{x}}} \cr & \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3x + \frac{3}{x} + 5}} \cr & \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3\left( {x - \frac{1}{x}} \right) + 5}} \cr & \Rightarrow x - \frac{1}{x} = 1{\text{ }} \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {x - \frac{1}{x}} \right)^3} = {\left( 1 \right)^3}{\text{ }} \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3\left( {x - \frac{1}{x}} \right) = 1 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3\left( 1 \right) = 1 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 4 \cr & \Rightarrow \frac{{{x^3} - \frac{1}{{{x^3}}}}}{{3\left( {x - \frac{1}{x}} \right) + 5}} \cr & \Rightarrow \frac{4}{{3 \times 1 + 5}} \cr & \Rightarrow \frac{4}{8} \cr & \Rightarrow \frac{1}{2} \cr} $$

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

$$\eqalign{ & a = 2 + \sqrt 3 \cr & \Rightarrow {a^2} = {\left( {2 + \sqrt 3 } \right)^2} \cr & \Rightarrow {a^2} = 4 + 3 + 4\sqrt 3 \cr & \Rightarrow {a^2} = 7 + 4\sqrt 3 \cr & \frac{1}{{{a^2}}} = \frac{1}{{7 + 4\sqrt 3 }} \cr & \Rightarrow \frac{1}{{{a^2}}} = \frac{{7 - 4\sqrt 3 }}{{\left( {7 + 4\sqrt 3 } \right)\left( {7 - 4\sqrt 3 } \right)}} \cr & \Rightarrow \frac{1}{{{a^2}}} = \frac{{7 - 4\sqrt 3 }}{1} \cr & \Rightarrow \frac{1}{{{a^2}}} = 7 - 4\sqrt 3 \cr & \therefore {a^2} + \frac{1}{{{a^2}}} \cr & = 7 + 4\sqrt 3 + 7 - 4\sqrt 3 \cr & = 14 \cr} $$

$$\eqalign{ & {x^2} + {y^2} + 1 = 2x \cr & {x^2} - 2x + 1 + {y^2} = 0 \cr & {\left( {x - 1} \right)^2} + {y^2} = 0 \cr & {\text{If }}{{\text{A}}^2} + {{\text{B}}^2} = 0 \cr} $$
[As powers are even it can possible only when A = 0 & B = 0]
$$\eqalign{ & \therefore x - 1 = 0 \cr & x = 1 \cr & y = 0 \cr & \therefore {x^3} + {y^5} \cr & = 1 + 0 \cr & = 1 \cr} $$

$$\eqalign{ & x\left( {x - 3} \right) = - 1 \cr & \Rightarrow \left( {x - 3} \right) = \frac{{ - 1}}{x} \cr & {\text{Taking cube on both sides}} \cr & \Rightarrow {\left( {x - 3} \right)^3} = {\left( {\frac{{ - 1}}{x}} \right)^3} \cr & \Rightarrow {x^3} - 27 - 9.x.\left( {x - 3} \right) = \frac{{ - 1}}{{{x^3}}} \cr & \Rightarrow {x^3} - 27 - 9 \times - 1 = \frac{{ - 1}}{{{x^3}}} \cr & \Rightarrow {x^3} - 27 + 9 = \frac{{ - 1}}{{{x^3}}} \cr & \Rightarrow {x^3} - 18 = \frac{{ - 1}}{{{x^3}}} \cr & \Rightarrow {x^3}\left( {{x^3} - 18} \right) = - 1 \cr} $$

$$\eqalign{ & {x^2} - 3x + 1 = 0 \cr & \Rightarrow {x^2} + 1 = 3x \cr & \Rightarrow x + \frac{1}{x} = 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 18 \cr & \therefore \frac{{{x^6} + {x^4} + {x^2} + 1}}{{{x^3}}}{\text{ }} \cr & = \frac{{{x^6}}}{{{x^3}}} + \frac{{{x^4}}}{{{x^3}}} + \frac{{{x^2}}}{{{x^3}}} + \frac{1}{{{x^3}}} \cr & = {x^3} + \frac{1}{{{x^3}}} + \frac{1}{x} + x \cr & = 18 + 3 \cr & = 21 \cr} $$

$$\eqalign{ & x + y = 15 \cr & \Rightarrow x - 10 = 5 - y \cr & \Rightarrow x - 10 = - \left( {y - 5} \right) \cr & {\text{Take cube on both sides}} \cr & \Rightarrow {\left( {x - 10} \right)^3} = - {\left( {y - 5} \right)^3} \cr & \Rightarrow {\left( {x - 10} \right)^3} + {\left( {y - 5} \right)^3} = 0 \cr} $$

$$\eqalign{ & a = \frac{{{b^2}}}{{b - a}} \cr & \Rightarrow a\left( {b - a} \right) = {b^2} \cr & \Rightarrow ab - {a^2} = {b^2} \cr & \Rightarrow {a^2} + {b^2} - ab = 0 \cr & \Rightarrow {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right) \cr & \Rightarrow {a^3} + {b^3} = 0 \cr} $$

$$\eqalign{ & 2x + 1 = 0 \cr & \Leftrightarrow x = \frac{{ - 1}}{2}{\text{ }} \cr & 3y - 9 = 0 \cr & \Leftrightarrow y = \frac{9}{3} = 3 \cr & \therefore \left( {x,y} \right) = \left( {\frac{{ - 1}}{2},3} \right) \cr & {\text{Point of intersection}} \cr} $$

$$\frac{1}{{{x^{99}}}}{\text{ + }}\frac{1}{{{x^{98}}}} + \frac{1}{{{x^{97}}}} + \frac{1}{{{x^{96}}}} + \frac{1}{{{x^{95}}}} + \frac{1}{{{x^{94}}}} + \frac{1}{x} - 1$$
  $$ = \frac{1}{{{{\left( { - 1} \right)}^{99}}}}{\text{ + }}\frac{1}{{{{\left( { - 1} \right)}^{98}}}} + \frac{1}{{{{\left( { - 1} \right)}^{97}}}} + \frac{1}{{{{\left( { - 1} \right)}^{96}}}} + $$         $$\frac{1}{{{{\left( { - 1} \right)}^{95}}}} + $$   $$\frac{1}{{{{\left( { - 1} \right)}^{94}}}} + $$   $$\frac{1}{{\left( { - 1} \right)}} - $$   $$1$$
$$\eqalign{ & = - 1 + 1 - 1 + 1 - 1 + 1 + \frac{1}{{ - 1}} - 1 \cr & = - 2 \cr} $$