$$\eqalign{ & {\text{Let }}\frac{a}{{q - r}} = \frac{b}{{r - p}} = \frac{c}{{p - q}} = k \cr & \frac{a}{{q - r}} = k{\text{ }}\left( {{\text{On multiplying by }}p} \right) \cr & pa = k\left( {pq - pr} \right)\,......(i) \cr & {\text{In the same way we can write}} \cr & {\text{qb = k}}\left( {qr - qp} \right)\,........(ii) \cr & {\text{And }}rc = k\left( {rp - rp} \right)\,.....iii) \cr & {\text{On adding equation (i), (ii) and (iii) }} \cr & pa + qb + rc \cr & = k\left( {pq - pr + qr - qp + rp - rq} \right) \cr & = 0 \cr} $$

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

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

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

$$\eqalign{ & {\text{We know that }} \cr & {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) \cr & {x^3} - p = \left( {x - 4} \right)\left( {{x^2} + 4x + 16} \right) \cr & \Rightarrow {x^3} - p = \left( {{x^3} - {4^3}} \right) \cr & \Rightarrow p = {4^3}{\text{ }}\left( {{\text{By comparison}}} \right) \cr & {\text{So, }}p = 64 \cr} $$

$$\eqalign{ & \frac{{5x}}{{4{x^2} + 10x + 1}} \cr & = \frac{5x}{{x\left( {4x + 10 + \frac{1}{x}} \right)}} \cr & = \frac{5}{{4x + \frac{1}{x} + 10}} \cr & = \frac{5}{{5 + 10}} \cr & = \frac{5}{{15}} \cr & = \frac{1}{3} \cr} $$

$$\eqalign{ & {\text{Third ratio,}} \cr & a:x::x:b \cr & {x^2} = ab \cr & \Rightarrow x = \sqrt {{{\left( {x - y} \right)}^2}{{\left( {{x^2} - {y^2}} \right)}^2}} \cr & \Rightarrow x = \left( {x - y} \right){\left( {{x^2} - {y^2}} \right)} \cr & \Rightarrow x = \left( {x - y} \right)\left( {x - y} \right)\left( {x + y} \right) \cr & \Rightarrow x = \left( {x + y} \right){\left( {x - y} \right)^2} \cr} $$

$$\eqalign{ & x + \sqrt 5 = 5 + \sqrt y \cr & {\text{Put , }}x = 5{\text{ and }}y = 5 \cr & 5 + \sqrt 5 = 5 + \sqrt 5 \cr & {\text{L}}{\text{.H}}{\text{.S}} = {\text{R}}{\text{.H}}{\text{.S}} \cr & \frac{{\sqrt x + y}}{{x + \sqrt y }} \cr & = \frac{{\sqrt 5 + 5}}{{5 + \sqrt 5 }} \cr & = 1 \cr} $$

$$\eqalign{ & pq\left( {p + q} \right) = 1 \cr & \Rightarrow p + q = \frac{1}{{pq}} \cr & \,\,\,\,\,\,\, \text{Cubing both side} \cr & \Rightarrow {p^3} + {q^3} +3 pq\left( {p + q} \right) = \frac{1}{{{p^3}{q^3}}} \cr & \,\,\,\,\,\,\, \text{Puting the value of } pq = \frac{1}{\left({p + q}\right)} \cr & \Rightarrow \frac{1}{{{p^3}{q^3}}} - {p^3} - {q^3} = \left( {\frac{{3}}{{p + q}}} \right)\left( {p + q} \right) \cr & \Rightarrow \frac{1}{{{p^3}{q^3}}} - {p^3} - {q^3} = 3 \cr} $$

$${p^3} - {q^3} = \left( {p - q} \right)\left\{ {{p^2} + {q^2} + pq} \right\}$$
$$ \Rightarrow \left( {p - q} \right)\left\{ {{{\left( {p + q} \right)}^2} - xpq} \right\} = $$       $$\left( {p - q} \right)$$ $$\left( {{p^2} + {q^2} + pq} \right)$$
$$\eqalign{ & \Rightarrow {p^2} + {q^2} + 2pq - xpq = {p^2} + {q^2} + pq \cr & \Rightarrow 2pq - pq = xpq \cr & \Rightarrow pq = xpq \cr & \Rightarrow x = 1 \cr} $$

$$\eqalign{ & \Rightarrow {p^3} - {q^3} = \left( {p - q} \right)\left\{ {{{\left( {p - q} \right)}^2} + xpq} \right\} \cr & \Rightarrow \left( {p - q} \right)\left\{ {{p^2} + {q^2} + pq} \right\} = \left( {p - q} \right)\left\{ {{p^2} + {q^2} - 2pq + xpq} \right\} \cr & \Rightarrow {p^2} + {q^2} + pq = {p^2} + {q^2} - 2pq + xpq \cr & \Rightarrow 3pq = xpq \cr & \Rightarrow x = 3 \cr} $$

$$\eqalign{ & {\text{Let the numbers are a, b}} \cr & {a^2} + {b^2} = 41 \cr & a + b = 9 \cr & \Rightarrow {\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab \cr & \Rightarrow {9^2} = 41 + 2ab \cr & \Rightarrow 81 - 41 = 2ab \cr & \Rightarrow ab = 20 \cr & {\text{Take }} \cr & a = 5 \cr & b = 4 \cr & \Rightarrow {a^3} + {b^3} = {5^3} + {4^3} \cr & \Rightarrow {a^3} + {b^3} = 125 + 64 \cr & \Rightarrow {a^3} + {b^3} = 189 \cr} $$

$$\eqalign{ & \left( {{x^4} + 64} \right) \cr & = {x^4} + {8^2} + 2.{x^2}.8 - 2.{x^2}.8 \cr & = {\left( {{x^2} + 8} \right)^2} - \left( {16{x^2}} \right) \cr & = {\left( {{x^2} + 8} \right)^2} - {\left( {4x} \right)^2} \cr & = \left( {{x^2} + 8 + 4x} \right)\left( {{x^2} + 8 - 4x} \right) \cr & = \left( {{x^2} + 4x + 8} \right)\left( {{x^2} - 4x + 8} \right) \cr} $$

$$\eqalign{ & a - b = 2{\text{ }} \cr & {a^3} - {b^3} = 26 \cr & \Rightarrow {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) \cr & \Rightarrow 26 = \left( 2 \right)\left( {{a^2} + ab + {b^2}} \right) \cr & \Rightarrow 13 = \left( {{a^2} + ab + {b^2}} \right)\,....(i) \cr & \Rightarrow 4 = 13 + ab \cr & \Rightarrow {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab \cr & \Rightarrow {\left( 2 \right)^2} = {a^2} + {b^2} + ab - 3ab \cr & \Rightarrow 3ab = 9 \cr & \Rightarrow ab = 3 \cr & \therefore {\left( {a + b} \right)^2} \cr & = {\left( {a - b} \right)^2} + 4ab \cr & = 4 + 4 \times 3 \cr & = 16 \cr} $$

$$\eqalign{ & \mathop {\mathop {{{\left( {x - 4} \right)}^3}}\limits_ \Downarrow }\limits_{{a^3}} + \mathop {\mathop {{{\left( {y - 2} \right)}^3}}\limits_ \Downarrow }\limits_{{b^3}} + \mathop {\mathop {{{\left( {z - 3} \right)}^3}}\limits_ \Downarrow }\limits_{{c^3} - 3abc} - 3\left( {x - 4} \right)\left( {y - 2} \right)\left( {z - 3} \right) \cr & \Rightarrow a + b + c = x - 4 + y - 2 + z - 3 \cr & \Rightarrow a + b + c = x + y + z - 9 \cr & \Rightarrow a + b + c = 9 - 9 \cr & \Rightarrow a + b + c = 0 \cr} $$
$${\text{So, }}{\left( {x - 4} \right)^3} + {\left( {y - 2} \right)^3} + $$     $${\left( {z - 3} \right)^3} - $$   $$3\left( {x - 4} \right)$$  $$\left( {y - 2} \right)$$ $$\left( {z - 3} \right)$$
$$ \Rightarrow 0$$

$$\eqalign{ & {\text{ }}x + \frac{1}{{9x}} = 4 \cr & {\text{Multiply by 3 both side}} \cr & \Rightarrow {\text{3}}x + \frac{1}{{3x}} = 12 \cr & {\text{Squaring both sides}} \cr & \Rightarrow {\text{9}}{x^2} + \frac{1}{{9{x^2}}} + 2 \times 3x \times \frac{1}{{3x}} = 144 \cr & \Rightarrow {\text{9}}{x^2} + \frac{1}{{9{x^2}}} + 2 = 144 \cr & \Rightarrow {\text{9}}{x^2} + \frac{1}{{9{x^2}}} = 142 \cr} $$

$$\eqalign{ & \frac{{{a^2} + {b^2}}}{{{c^2}}} = \frac{{{b^2} + {c^2}}}{{{a^2}}} = \frac{{{c^2} + {a^2}}}{{{b^2}}} = \frac{1}{k} \cr & {\text{Put }}a = b = c = 1 \cr & \Rightarrow \frac{{1 + 1}}{1} + \frac{{1 + 1}}{1} + \frac{{1 + 1}}{1} = \frac{1}{k} \cr & \Rightarrow 2 = 2 = 2 = \frac{1}{k} \cr & \Rightarrow k = \frac{1}{2} \cr} $$

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

$$\eqalign{ & c + \frac{1}{c} = \sqrt 3 \cr & {\text{On cubing both side}} \cr & \Rightarrow {\left( {c + \frac{1}{c}} \right)^3} = 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} + 3.c.\frac{1}{c}\left( {c + \frac{1}{c}} \right) = 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {c^3} + \frac{1}{{{c^3}}} = 0 \cr} $$

$${x^3} + {y^3} + {z^3} - 3xyz$$
$$ = \frac{1}{2}\left( {x + y + z} \right)$$  $$\left[ {{{\left( {x - y} \right)}^2} + {{\left( {y - z} \right)}^2} + {{\left( {z - x} \right)}^2}} \right]$$
$$ = \frac{1}{2}\left( {222 + 223 + 225} \right)$$    $$\left[ {{{\left( {222 - 223} \right)}^2} + {{\left( {223 - 225} \right)}^2} + {{\left( {225 - 222} \right)}^2}} \right]$$
$$\eqalign{ & = \frac{1}{2}\left( {670} \right)\left( {1 + 4 + 9} \right) \cr & = \frac{1}{2} \times 670 \times 14 \cr & = 4690 \cr} $$

This is possible only when
$$\eqalign{ & {\left( {x - 3} \right)^2} = 0 \cr & x = 3 \cr & {\left( {y - 4} \right)^2} = 0 \cr & y = 4 \cr & {\left( {z - 5} \right)^2} = 0 \cr & z = 5 \cr & {\text{Then, }}\left( {x + y + z} \right) \cr & = 3 + 4 + 5 \cr & = 12 \cr} $$

$$\eqalign{ & 2x + \frac{1}{{3x}} = 5 \cr & 6{x^2}{\text{ + 1 = 15x}}\,......{\text{(i)}} \cr & {\text{Now,}}\frac{{5x}}{{6{x^2} + 20x + 1}} \cr & = \frac{{5x}}{{6{x^2} + 1 + 20x}} \cr & \left[ {{\text{From equation (i)}}} \right] \cr & = \frac{{5x}}{{15x + 20x}}{\text{ }} \cr & = \frac{{5x}}{{35x}} \cr & = \frac{1}{7} \cr} $$

$$\eqalign{ & x = {\left( {0.25} \right)^{\frac{1}{2}}} \cr & x = \root 2 \of {0.25} = 0.5{\text{ }} \cr & y = {\left( {0.4} \right)^2} \cr & y = 0.16 \cr & z = {\left( {0.216} \right)^{\frac{1}{2}}} \cr & z = \root 2 \of {0.216} = 0.464 \cr & \therefore x > z > y \cr} $$

$$\eqalign{ & a\left( {x + y} \right) = b\left( {x - y} \right) = 2ab \cr & x + y = 2b \cr & {\text{On squraing both side}} \cr & \Rightarrow {x^2} + {y^2} + 2xy = 4{b^2}\,.....(i) \cr & \Rightarrow b\left( {x - y} \right) = 2ab \cr & \Rightarrow x - y = 2a \cr & {\text{On squraing both side}} \cr & \Rightarrow {x^2} + {y^2} - 2xy = 4{a^2}\,.....(ii) \cr & {\text{Add equation (i) and (ii)}} \cr & \Rightarrow 2\left( {{x^2} + {y^2}} \right) = 4\left( {{a^2} + {b^2}} \right) \cr} $$

$$\eqalign{ & {\text{Let numbers are }} \cr & x = 2r,{\text{ }}y = 3r,{\text{ }}z = 5r \cr & {\text{Sum, }}x + y + z = 2r + 3r + 5r \cr & \Rightarrow x + y + z = 10r \cr & \Rightarrow x + y + z = 80 \cr & r = 8,x = 16,y = 24,z = 40 \cr & {\text{Then, }}z = ax - 8 \cr & \Rightarrow 40 = a \times 16 - 8 \cr & \Rightarrow a = 3 \cr} $$

$$\eqalign{ & \left( {x - 2} \right)\left( {x - p} \right) = {x^2} - ax + 6 \cr & {x^2} - \left( {2 + p} \right)x + 2p = {x^2} - ax + 6 \cr & {\text{Comparision the cofficients}} \cr & 2 + p = a \cr & 2p = 6 \cr & \Leftrightarrow p = 3 \cr & 2 + 3 = a \cr & \Leftrightarrow a = 5 \cr & {\text{Then , }} \cr & p = 3,{\text{ }}a = 5 \cr & a - p = 5 - 3 \cr & \Leftrightarrow a - p = 2 \cr} $$

$$\eqalign{ & x = a + \frac{1}{a} \cr & {x^2} = {a^2} + \frac{1}{{{a^2}}} + 2 \cr & {y^2} = {a^2} + \frac{1}{{{a^2}}} - 2 \cr & {\text{Now, }} \cr & {x^4} + {y^4} - 2{x^2}{y^2} \cr & = {\left( {{x^2} - {y^2}} \right)^2} \cr & = {\left( {{a^2} + \frac{1}{{{a^2}}} + 2 - {a^2} - \frac{1}{{{a^2}}} + 2} \right)^2} \cr & = {\left( 4 \right)^2} \cr & = 16 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = \sqrt 3 \cr & {\text{Cubing both side}} \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3.x.\frac{1}{x}\left( {x + \frac{1}{x}} \right) = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\left( {\sqrt 3 } \right) = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 3\sqrt 3 - 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 0 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 5{\text{ then, }}\frac{{5x}}{{{x^2} + 5x + 1}} \cr & \Rightarrow \frac{5}{{x + 5 + \frac{1}{x}}} \cr & \Rightarrow \frac{5}{{x + \frac{1}{x} + 5}} \cr & \Rightarrow \frac{5}{{5 + 5}} \cr & \Rightarrow \frac{1}{2} \cr} $$

$$\eqalign{ & \frac{1}{a}\left( {{a^2} + 1} \right) = 3 \cr & \Rightarrow a + \frac{1}{a} = 3 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3.a.\frac{1}{a}\left( {a + \frac{1}{a}} \right) = {3^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} + 3\left( 3 \right) = {3^3} \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 27 - 9 \cr & \Rightarrow {a^3} + \frac{1}{{{a^3}}} = 18 \cr & \Rightarrow \frac{{{a^6} + 1}}{{{a^3}}} = 18 \cr} $$