$$\eqalign{ & {\text{8}} \times {\text{2}} \cr & = 8 + 2 + \frac{8}{2} \cr & = 10 + 4 \cr & = 14 \cr} $$

$$\eqalign{ & {t^2} - 4t + 1 = 0 \cr & \Rightarrow {t^2} + 1 = 4t \cr & \Rightarrow \frac{{{t^2} + 1}}{t} = \frac{{4t}}{t} \cr & \Rightarrow t + \frac{1}{t} = 4 \cr & \left[ {{\text{Take cube both sides}}} \right] \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} + 3.t.\frac{1}{t}\left( {t + \frac{1}{t}} \right) = 64 \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} + 3\left( 4 \right) = 64 \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} = 64 - 12 \cr & \Rightarrow {t^3} + \frac{1}{{{t^3}}} = 52 \cr} $$

$$\eqalign{ & \root 3 \of a + \root 3 \of b + \root 3 \of c \cr & {\text{Take cube both sides}} \cr & {\left( {\root 3 \of a + \root 3 \of b } \right)^3} = {\left( {\root 3 \of c } \right)^3} \cr & \Rightarrow a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}\left( {\root 3 \of a + \root 3 \of b } \right) = c \cr & \Rightarrow a + b + 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}{c^{\frac{1}{3}}} = c \cr & \Rightarrow a + b - c = - 3{a^{\frac{1}{3}}}{b^{\frac{1}{3}}}{c^{\frac{1}{3}}} \cr & {\text{Again take cube both sides}} \cr & \Rightarrow {\left( {a + b - c} \right)^3} = - 27abc \cr & \Rightarrow {\left( {a + b - c} \right)^3} + 27abc = 0 \cr & \cr & {\bf{Alternate:}} \cr & {\text{Put value of }} \cr & a = 0 \cr & b = 1 \cr & c = 1 \cr & {\text{Value of }}{\left( {a + b - c} \right)^3} + 27abc \cr & = {\left( {0 + 1 - 1} \right)^3} + 27 \times 0 \times 1 \times 1 \cr & = 0 \cr} $$

$$\eqalign{ & {a^2} - 4a - 1 = 0 \cr & {a^2} - 1 = 4a \cr & a - \frac{1}{a} = 4 \cr & {\text{Squring both sides}} \cr & {a^2} + \frac{1}{{{a^2}}} - 2 = 16 \cr & \Rightarrow {a^2} + \frac{1}{{{a^2}}} = 18 \cr & \therefore {a^2} + \frac{1}{{{a^2}}} + 3a - \frac{3}{a} \cr & = {a^2} + \frac{1}{{{a^2}}} + 3\left( {a - \frac{1}{a}} \right) \cr & = 18 + 3 \times 4 \cr & = 18 + 12 \cr & = 30 \cr} $$

$$\eqalign{ & {\text{ }}\frac{a}{b} = \frac{7}{9},{\text{ }}\frac{b}{c} = \frac{3}{5} \cr & \therefore a\,\,\,:\,\,\,b\,\,\,:\,\,\,c \cr & \,\,\,\,\,7\,\,\,:\,\,\,9 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{3_{ \times 3}}:\,\,\,{5_{ \times 3}} \cr & \overline {{\text{ }}\,\,{\text{ }}7\,\,\,:\,\,\,\,9\,\,:\,\,15{\text{ }}} \cr} $$

$$\eqalign{ & a \oplus b = \left( {a \times b} \right) + b \cr & \Rightarrow 5 \oplus 7 \cr & = \left( {5 \times 7} \right) + 7 \cr & = 35 + 7 \cr & = 42 \cr} $$

$$\eqalign{ & {10^{0.48}} = x{\text{ }} \cr & {10^{0.70}} = y{\text{ }} \cr & {\text{and }}{x^z} = {y^2} \cr & \therefore {\left( {{{10}^{0.48}}} \right)^z} = {\left( {{{10}^{0.70}}} \right)^2} \cr & \Rightarrow {10^{0.48z}} = {10^{1.40}} \cr} $$
If a^x = a^y, if base equal power are equal : (x = y)
$$\eqalign{ & \therefore 0.48z = 1.40 \cr & \Rightarrow z = \frac{{140}}{{48}} \cr & \Rightarrow z = \frac{{35}}{{12}} \cr & \Rightarrow z = 2.9 \cr} $$

$$\eqalign{ & {\text{A}}:{\text{B}} = \frac{1}{2}:\frac{3}{8} \cr & \Rightarrow {\text{A}}:{\text{B}} = 8:6 \cr & \Rightarrow {\text{A}}:{\text{B}} = 4:3 \cr & {\text{B}}:{\text{C}} = \frac{1}{3}:\frac{5}{9} \cr & \Rightarrow {\text{B}}:{\text{C}} = 9:15 \cr & \Rightarrow {\text{B}}:{\text{C}} = 3:5 \cr & {\text{C}}:{\text{D}} = \frac{5}{6}:\frac{3}{4} \cr & \Rightarrow {\text{C}}:{\text{D}} = 20:18 \cr & \Rightarrow {\text{C}}:{\text{D}} = 10:9 \cr & {\text{A}}:{\text{ B }}:{\text{C }}:{\text{D}} \cr & 4{\text{ }}:{\text{ }}3 \cr & \,\,\,\,\,\,\,\,\,\,3{\text{ }}:{\text{ }}5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,10{\text{ }}:{\text{ }}9 \cr & \overline {\underline {{\text{ }}8{\text{ }}:6{\text{ }}:10{\text{ }}:{\text{ }}9{\text{ }}} } \cr} $$

$$\eqalign{ & {\text{A}}:{\text{B}}:{\text{C}} \cr & {\text{ }}2:\,\,3\,\,:4 \cr & \therefore \frac{{\text{A}}}{{\text{B}}}:\frac{{\text{B}}}{{\text{C}}}:\frac{{\text{C}}}{{\text{A}}} \cr & \left( {{\text{multiply with ABC}}} \right) \cr & \therefore \frac{{{\text{A}} \times {\text{ABC}}}}{{\text{B}}}{\text{:}}\frac{{{\text{B}} \times {\text{ABC}}}}{{\text{C}}}{\text{:}}\frac{{{\text{C}} \times {\text{ABC}}}}{{\text{A}}} \cr & \Rightarrow {{\text{A}}^2}{\text{C}}:{{\text{B}}^2}{\text{A}}:{\text{B}}{{\text{C}}^2} \cr & \Rightarrow {\left( 2 \right)^2} \times 4:{\left( 3 \right)^2} \times 2:{\left( 4 \right)^2} \times 3 \cr & \Rightarrow 16:18:48 \cr & \Rightarrow 8:9:24 \cr & \cr & {\bf{Alternate:}} \cr & \frac{{\text{A}}}{{\text{B}}}:\frac{{\text{B}}}{{\text{C}}}:\frac{{\text{C}}}{{\text{A}}} = \frac{2}{3}:\frac{3}{4}:\frac{4}{2} \cr & \Rightarrow \frac{2}{3} \times 12:\frac{3}{4} \times 12:\frac{4}{2} \times 12 \cr & \Rightarrow 8:9:24 \cr} $$

$$\left( {{\text{1 + }}\frac{1}{x}} \right)$$ $$\left( {{\text{1 + }}\frac{1}{{x + 1}}} \right)$$  $$\left( {{\text{1 + }}\frac{1}{{x + 2}}} \right)$$  $$\left( {{\text{1 + }}\frac{1}{{x + 3}}} \right)$$
Taking L.C.M of each term
$$ \Rightarrow \left( {\frac{{x + 1}}{x}} \right)$$ $$\left( {\frac{{x + 1 + 1}}{{x + 1}}} \right)$$  $$\left( {\frac{{x + 2 + 1}}{{x + 2}}} \right)$$  $$\left( {\frac{{x + 3 + 1}}{{x + 3}}} \right)$$
$$\eqalign{ & \Rightarrow \frac{1}{x} \times \left( {x + 4} \right) \cr & \Rightarrow \frac{{x + 4}}{x} \cr} $$

$$\eqalign{ & \frac{a}{b}{\text{ = }}\frac{2}{3}{\text{ and }}\frac{b}{c}{\text{ = }}\frac{4}{5}\,\,\left( {{\text{Given}}} \right) \cr & or\,\frac{c}{b} = \frac{5}{4} \cr & \frac{{a + b}}{{b + c}} \cr & = \frac{{b\left( {\frac{a}{b} + 1} \right)}}{{b\left( {\frac{c}{b} + 1} \right)}} \cr & = \frac{{\frac{a}{b} + 1}}{{\frac{c}{b} + 1}} \cr & = \frac{{\left( {\frac{2}{3} + 1} \right)}}{{\left( {\frac{5}{4} + 1} \right)}} \cr & = \frac{{\frac{2 + 3}{3}}}{{\frac{{5 + 4}}{4}}} \cr & = \frac{{5 \times 4}}{{3 \times 9}} \cr & = \frac{{20}}{{27}} \cr & \therefore \frac{{a + b}}{{b + c}} = \frac{{20}}{{27}} \cr & {\bf{Alternate:}} \cr & a{\text{ }}\,\,\,{\text{ }}\,{\text{ }}\,\,\,\,{\text{ }}b{\text{ }}\,\,\,\,{\text{ }}\,\,\,\,{\text{ }}\,\,\,{\text{ }}c \cr & {2_{ \times \left( 4 \right)}}\,\,\,\,\,{\text{ }}{3_{ \times \left( 4 \right)}} \cr & \,{\text{ }}\,{\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,{4_{ \times \left( 3 \right)}}{\text{ }}\,\,\,\,\,\,{5_{ \times \left( 3 \right)}} \cr & \overline {\underline {8{\text{ }}\,\,\,\,\,\,\,\,\,\,\,{\text{ }}12{\text{ }}\,\,\,\,\,\,\,\,\,\,{\text{ }}15\,\,\,\,} } \cr & \therefore \frac{{a + b}}{{b + c}} = \frac{{8 + 12}}{{12 + 15}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20}}{{27}} \cr} $$

$$\eqalign{ & \frac{{2a + b}}{{a + 4b}} = 3{\text{ }}\left( {{\text{Given}}} \right) \cr & \Rightarrow 2a + b = 3\left( {a + 4b} \right) \cr & \Rightarrow 2a + b = 3a + 12b \cr & \Rightarrow - a = 11b \cr & \Rightarrow a = - 11b \cr & \therefore \frac{{a + b}}{{a + 2b}} \cr & \Rightarrow \frac{{ - 11b + b}}{{ - 11b + 2b}} \cr & \Rightarrow \frac{{ - 10b}}{{ - 9b}} \cr & \Rightarrow \frac{{10}}{9} \cr} $$

$$\eqalign{ & y:x = 4:15 \cr & \therefore \frac{y}{x} = \frac{4}{{15}} \cr & \therefore \frac{{x - y}}{{x + y}} \cr & \Rightarrow \frac{{x\left( {1 - \frac{y}{x}} \right)}}{{x\left( {1 + \frac{y}{x}} \right)}} \cr & {\text{Taking }}x{\text{ common}} \cr & \Rightarrow \frac{{1 - \frac{4}{{15}}}}{{1 + \frac{4}{{15}}}} \cr & \Rightarrow \frac{{11}}{{15}} \times \frac{{15}}{{19}} \cr & \Rightarrow \frac{{11}}{{19}} \cr} $$

$$\eqalign{ & {\left( {x - 3} \right)^2}{\text{ + }}{\left( {y - 5} \right)^2}{\text{ + }}{\left( {z - 4} \right)^2} = 0 \cr & \therefore {\left( {x - 3} \right)^2} = 0{\text{ }}x = 3 \cr & {\left( {y - 5} \right)^2} = 0{\text{ }}y = 5 \cr & {\left( {z - 4} \right)^2} = 0{\text{ }}z = 4{\text{ }} \cr & \therefore \frac{{{x^2}}}{9}{\text{ + }}\frac{{{y^2}}}{{25}}{\text{ + }}\frac{{{z^2}}}{{16}} \cr & \Rightarrow \frac{9}{9} + \frac{{25}}{{25}} + \frac{{16}}{{16}} \cr & \Rightarrow 3 \cr} $$

 $${\text{Given}}\,x = \root 3 \of {a + \sqrt {{a^2} + {b^3}} } \,\,+ $$      $$\root 3 \of {a - \sqrt {{a^2} + {b^3}} } $$
$$\eqalign{ & {\text{Let}}\,z = \sqrt {{a^2} + {b^3}} \cr & \therefore x = \root 3 \of {a + z} + \root 3 \of {a - z} \cr & {\text{Cubing both side}} \cr} $$
  $$\therefore {x^3} = {\left( {\root 3 \of {a + z} } \right)^3} + {\left( {\root 3 \of {a - z} } \right)^3} \, + $$      $$\,3{\left( {a + z} \right)^{\frac{1}{3}}}$$  $${\left( {a - z} \right)^{\frac{1}{3}}} \times\, $$   $$\left( {\root 3 \of {a + z} + \root 3 \of {a - z} } \right)$$
 $$ \,\Rightarrow {x^3} = a + z + a - z \,+\, $$    $$3{\left( {{a^2} + az - az - {z^2}} \right)^{\frac{1}{3}}}$$    $$ \times \left( x \right)$$
$$\eqalign{ & \Rightarrow {x^3} = 2a + 3{\left( {{a^2} - {z^2}} \right)^{\frac{1}{3}}} \times \left( x \right) \cr & {\text{Put the value of }}z \cr & \Rightarrow {x^3} = 2a + 3{\left( {{a^2} - {a^2} - {b^3}} \right)^{\frac{1}{3}}} \times \left( x \right) \cr & \Rightarrow {x^3} = 2a + 3{\left( { - {b^3}} \right)^{\frac{1}{3}}} \times \left( x \right) \cr & \Rightarrow {x^3} = 2a - 3bx \cr & \therefore {x^3} + 3bx = 2a \cr} $$

$$\eqalign{ & \frac{{{x^{24}} + 1}}{{{x^{12}}}} = 7{\text{ }}\left( {{\text{Given}}} \right) \cr & \Rightarrow \frac{{{x^{24}}}}{{{x^{12}}}} + \frac{1}{{{x^{12}}}} = 7 \cr & \Rightarrow {x^{12}} + \frac{1}{{{x^{12}}}} = 7 \cr & {\text{Cubing both sides}} \cr & \Rightarrow {\left( {{x^{12}} + \frac{1}{{{x^{12}}}}} \right)^3} = {7^3} \cr} $$
  $$ \Rightarrow {x^{36}} + \frac{1}{{{x^{36}}}} + \frac{{3 \times {x^{12}} \times 1}}{{{x^{12}}}}$$     $$\left( {{x^{12}} + \frac{1}{{{x^{12}}}}} \right) = $$    $$343$$
$$\eqalign{ & \Rightarrow {x^{36}} + \frac{1}{{{x^{36}}}} + 3\left( 7 \right) = 343 \cr & \Rightarrow {x^{36}} + \frac{1}{{{x^{36}}}} = 343 - 21 \cr & \Rightarrow {x^{36}} + \frac{1}{{{x^{36}}}} = 322 \cr & \Rightarrow \frac{{{x^{72}} + 1}}{{{x^{36}}}} = 322 \cr} $$

$$\eqalign{ & {\text{47}}{\text{.2506}} = {\text{4A}} + {\text{7B}} + {\text{2C}} + \frac{5}{{\text{D}}} + {\text{6E}} \cr }$$
⇒ 47.2506 = 4 × 10 + 7 × 1 + 2 × 0.1000 + 5 × 0.0100 + 0 + 6 × 0.0001
$$\eqalign{ & \therefore {\text{A}} = 10, \cr & {\text{B}} = 1, \cr & {\text{C}} = 0.1000, \cr & {\text{D}} = \frac{1}{{\frac{1}{{100}}}} = 100 \cr & {\text{E}} = 0.0001 \cr & \therefore {\text{5A}} + {\text{3B}} + {\text{6C}} + {\text{D}} + {\text{3E}} \cr & = 5 \times 10 + 3 \times 1 + 6 \times 0.1 + 100 + 3 \times 0.0001 \cr & = 50 + 3 + 0.6 + 100 + 0.0003 \cr & = 153.6003 \cr} $$

$$\eqalign{ & x*y = {x^2} + {y^2} - xy \cr & {\text{ }}9*11 = {\left( 9 \right)^2} + {\left( {11} \right)^2} - 11 \times 9 \cr & \Rightarrow 9*11 = 81 + 121 - 99 \cr & \Rightarrow 9*11 = 103 \cr} $$

$$\eqalign{ & x + \frac{1}{x} = 3 \cr & \left( {{\text{Squaring both sides}}} \right) \cr & {x^2} + \frac{1}{{{x^2}}} = 7 \cr & {\text{On cubing both sides}} \cr & {x^3} + \frac{1}{{{x^3}}} + 3.x.\frac{1}{x}\left( {x + \frac{1}{x}} \right) = 27 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3 \times 3 = 27 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 18 \cr & \therefore \left( {{x^3} + \frac{1}{{{x^3}}}} \right)\left( {{x^2} + \frac{1}{{{x^2}}}} \right) = 18 \times 7 \cr & \Rightarrow \left( {{x^5} + \frac{1}{{{x^5}}}} \right) + \left( {x + \frac{1}{x}} \right) = 126 \cr & \Rightarrow \left( {{x^5} + \frac{1}{{{x^5}}}} \right) + 3 = 126 \cr & \Rightarrow \left( {{x^5} + \frac{1}{{{x^5}}}} \right) = 123 \cr} $$

$$\eqalign{ & {\text{ }}x + \frac{1}{x} = 2{\text{, }}\,\,\,x \ne 0 \cr & {\text{Put }}x = 1 \cr & 1 + 1 = 2 \cr & \therefore {x^2}{\text{ + }}\frac{1}{{{x^2}}} \cr & = 1 + 1 \cr & = 2 \cr} $$

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

$$\eqalign{ & p + q = 10\,.........{\text{(i)}} \cr & pq = 5 \cr & {\text{Squaring both sides of equation (i)}} \cr & {\left( {p + q} \right)^2} = {\left( {10} \right)^2} \cr & {p^2} + {q^2} + 2pq = 100 \cr & {p^2} + {q^2} + 2 \times 5 = 100 \cr & {p^2} + {q^2} = 90 \cr & {\text{Now,}} \cr & \therefore \frac{p}{q}{\text{ + }}\frac{q}{p} \cr & = \frac{{{p^2} + {q^2}}}{{pq}} \cr & = \frac{{90}}{5} \cr & = 18 \cr} $$

$$\eqalign{ & x \propto \frac{1}{{{y^2}}} \cr & \left( {{\text{Inversely proportional}}} \right) \cr & x = \frac{k}{{{y^2}}} \cr & \left( {{\text{Given}}} \right), \cr & \left( {y = 2} \right){\text{ for }}\left( {x = 1} \right) \cr & \therefore 1 = \frac{k}{{{{\left( 2 \right)}^2}}} \cr & \Rightarrow 1 = \frac{k}{4} \cr & \Rightarrow k = 4 \cr & \therefore {\text{For }}y = 6 \cr & x = \frac{4}{{{{\left( 6 \right)}^2}}} \cr & x = \frac{4}{{36}} \cr & x = \frac{1}{9} \cr} $$

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

$$\eqalign{ & {\text{Given, }}x = 2 \cr & {\text{Find }}{x^3} + 27{x^2} + 243x + 631 \cr & {\text{To put value }}x = 2{\text{ }} \cr & \Rightarrow {2^3} + 27 \times {2^2} + \left( {243 \times 2} \right) + 631 \cr & \Rightarrow 8 + 108 + 486 + 631 \cr & \Rightarrow 1233 \cr} $$

$$\eqalign{ & x - \frac{1}{x} = 3 \cr & \left( {{\text{By cubing both sides}}} \right) \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3.x.\frac{1}{x}\left( {x - \frac{1}{x}} \right) = 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} - 3.\left( 3 \right) = 27 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 27 + 9 \cr & \Rightarrow {x^3} - \frac{1}{{{x^3}}} = 36 \cr} $$

$$\eqalign{ & x + y + z = 6 \cr & {\left( {x - 1} \right)^3}{\text{ + }}{\left( {y - 2} \right)^3}{\text{ + }}{\left( {z - 3} \right)^3} \cr & \therefore {\text{As, }}x + y + z = 6 \cr & {\text{Take values}} \cr & x = 1 \cr & y = 2 \cr & z = 3 \cr & \left( {1 + 2 + 3} \right) = 6 \cr & \therefore {\left( {1 - 1} \right)^3}{\text{ + }}{\left( {2 - 2} \right)^3}{\text{ + }}{\left( {3 - 3} \right)^3} \cr & = 0 \cr} $$
Now assume values in options.
Option 'D' satisfies the given relation.
Hence 'D' is correct.

$$\eqalign{ & {\left( {x + \frac{1}{x}} \right)^2} = 3 \cr & \Rightarrow x + \frac{1}{x} = \sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} + 3\sqrt 3 = 3\sqrt 3 \cr & \Rightarrow {x^3} + \frac{1}{{{x^3}}} = 0 \cr & \Rightarrow {x^6} + 1 = 0 \cr & \Rightarrow {x^6} = - 1 \cr & \Rightarrow {x^{72}} + {x^{66}} + {x^{54}} + {x^{24}} + {x^6} + 1 \cr & \Rightarrow {\left( {{x^6}} \right)^{12}} + {\left( {{x^6}} \right)^{11}} + {\left( {{x^6}} \right)^9} + {\left( {{x^6}} \right)^4} + {x^6} + 1 \cr & \Rightarrow {\left( { - 1} \right)^{12}} + {\left( { - 1} \right)^{11}} + {\left( { - 1} \right)^9} + {\left( { - 1} \right)^4} - 1 + 1 \cr & \Rightarrow 1 - 1 - 1 + 1 - 1 + 1 \cr & \Rightarrow 0 \cr} $$

$$\frac{{4 + 3\sqrt 3 }}{{7 + 4\sqrt 3 }}$$
(By rationalization of denominator)
$$\eqalign{ & = \frac{{4 + 3\sqrt 3 }}{{7 + 4\sqrt 3 }} \times \frac{{7 - 4\sqrt 3 }}{{7 - 4\sqrt 3 }} \cr & = \frac{{\left( {4 + 3\sqrt 3 } \right)\left( {7 - 4\sqrt 3 } \right)}}{{49 - 48}} \cr} $$

$$\eqalign{ & {a^3} - {b^3} = 56 \cr & \Rightarrow a - b = 2 \cr & \,\,\,\,\left( {{\text{By cubing}}} \right) \cr & \Rightarrow {a^3} - {b^3} - 3ab\left( {a - b} \right) = {\left( 2 \right)^2} \cr & \Rightarrow 56 - 3ab \times 2 = 8 \cr & \Rightarrow - 6ab = 8 - 56 \cr & \Rightarrow 6ab = 48 \cr & \Rightarrow ab = 8 \cr & \left( {a - b} \right) = 2 \cr & \,\,{\text{ }}\left( {{\text{By squaring}}} \right) \cr & \Rightarrow {\left( {a - b} \right)^2} = {\left( 2 \right)^2} \cr & \Rightarrow {a^2} + {b^2} - 2ab = 4 \cr & \Rightarrow {a^2} + {b^2} = 4 + 2ab \cr & \Rightarrow {a^2} + {b^2} = 4 + 2 \times 8 \cr & \Rightarrow {a^2} + {b^2} = 20 \cr} $$