According to question,

Factors are :
3 × 3 × 3 × 2 × 2 × 2 × 2 × 2 × 5
∴ It must be multiplied by
= 2 × 5 × 5
= 50

$$\eqalign{ & x = \sqrt 3 + \sqrt 2 \cr & \frac{1}{x} = \frac{1}{{\sqrt 3 + \sqrt 2 }} \times \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \cr & \frac{1}{x} = \sqrt 3 - \sqrt 2 \cr & {x^3} - \frac{1}{{{x^3}}} \cr & = {\left[ {x - \frac{1}{x}} \right]^3} + 3 \times x \times \frac{1}{x}\left( {x - \frac{1}{x}} \right) \cr} $$
  $$ = {\left( {\sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 } \right)^3} + $$      $$3\left( {\sqrt 3 + \sqrt 2 - \sqrt 3 + \sqrt 2 } \right)$$
$$\eqalign{ & = {\left( {2\sqrt 2 } \right)^3} + 3\left( {2\sqrt 2 } \right) \cr & = 16\sqrt 2 + 6\sqrt 2 \cr & = 22\sqrt 2 \cr} $$

($$\sqrt {6} $$  + $$\sqrt {10} $$  - $$\sqrt {21} $$  - $$\sqrt {35} $$) × ($$\sqrt {6} $$  - $$\sqrt {10} $$  + $$\sqrt {21} $$  - $$\sqrt {35} $$)
= {($$\sqrt {6} $$  - $$\sqrt {35} $$) + ($$\sqrt {10} $$  - $$\sqrt {21} $$)} × {($$\sqrt {6} $$  - $$\sqrt {35} $$) - ($$\sqrt {10} $$  - $$\sqrt {21} $$)}
= ($$\sqrt {6} $$  - $$\sqrt {35} $$)² - ($$\sqrt {10} $$  - $$\sqrt {21} $$)²
= 6 + 35 - 2$$\sqrt {210} $$  - 10 - 21 + 2$$\sqrt {210} $$
= 41 - 31
= 10

$$\eqalign{ & 2ab \cr & = \left( {{a^2} + {b^2}} \right) - {\left( {a - b} \right)^2} \cr & = 29 - 9 = 20 \cr & \therefore ab = 10 \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \,\,\frac{{\sqrt {24} + \sqrt {216} }}{{\sqrt {96} }} \cr & = \frac{{\sqrt {2 \times 2 \times 6} + \sqrt {6 \times 6 \times 6} }}{{\sqrt {4 \times 4 \times 6} }} \cr & = \frac{{2\sqrt 6 + 6\sqrt 6 }}{{4\sqrt 6 }} \cr & = \frac{8}{4} \cr & = 2 \cr} $$

Let the number of men in the 1st, 2nd, 3rd and 4th battalions be x, y, z and t respectively.
Then,
$$\eqalign{ & \frac{1}{2}x = \frac{2}{3}y = \frac{3}{4}z = \frac{4}{5}t \cr & \Rightarrow x = \frac{4}{3}y, \cr & \,\,\,\,\,\,\,\,z = \frac{8}{9}y, \cr & \,\,\,\,\,\,\,\,\,t = \frac{5}{6}y{\text{ }} \cr & {\text{Now,}} \cr & x + y + z + t = 7300 \cr & \Rightarrow \frac{4}{3}y + y + \frac{8}{9}y + \frac{5}{6}y = 7300 \cr & \Rightarrow \frac{{24y + 18y + 16y + 15y}}{{18}} = 7300 \cr & \Rightarrow 73y = 7300 \times 18 \cr & \Rightarrow y = 1800 \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \,\,\,\left[ {4 - \frac{5}{{1 + \frac{1}{{3 + \frac{1}{{2 + \frac{1}{4}}}}}}}} \right] \cr & = 4 - \frac{5}{{1 + \frac{1}{{3 + \frac{1}{{\frac{9}{4}}}}}}} \cr & = 4 - \frac{5}{{1 + \frac{1}{{3 + \frac{4}{9}}}}} \cr & = 4 - \frac{5}{{1 + \frac{1}{{\frac{{31}}{9}}}}} \cr & = 4 - \frac{5}{{1 + \frac{9}{{31}}}} \cr & = 4 - \frac{5}{{\frac{{40}}{{31}}}} \cr & = 4 - \frac{{5 \times 31}}{{40}} \cr & = 4 - \frac{{31}}{8} \cr & = \frac{1}{8} \cr & {\text{According to question}} \cr & \frac{1}{8}{\text{part = 10 minutes}} \cr & {\text{1 part = 10 minutes}} \cr & \frac{3}{5}{\text{ part = 80}} \times \frac{3}{5} \cr & {\text{ = 48 minutes}} \cr} $$

Let the number of students in a row be 'x'
According to question
⇒ x × x = 1369
⇒ x² = 1369
⇒ x = 37

$$\eqalign{ & {\text{Since }}\left( {x - y} \right) + \left( {y - z} \right) + \left( {z - x} \right) = 0 \cr & {\text{So,}}{\left( {x - y} \right)^3} + {\left( {y - z} \right)^3} + {\left( {z - x} \right)^3} \cr & = 3\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right) \cr & {\text{Since}}\left( {{x^2} - {y^2}} \right) + \left( {{y^2} - {z^2}} \right) + \left( {{z^2} - {x^2}} \right) = 0 \cr & {\text{So,}}\left( {{x^2} - {y^2}} \right) + \left( {{y^2} - {z^2}} \right) + \left( {{z^2} - {x^2}} \right) \cr & = 3\left( {{x^2} - {y^2}} \right)\left( {{y^2} - {z^2}} \right)\left( {{z^2} - {x^2}} \right) \cr & \therefore {\text{Given expression }} \cr & {\text{ = }}\frac{{3\left( {x - y} \right)\left( {y - z} \right)\left( {z - x} \right)}}{{3\left( {{x^2} - {y^2}} \right)\left( {{y^2} - {z^2}} \right)\left( {{z^2} - {x^2}} \right)}} \cr & = \frac{1}{{\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)}} \cr & = {\left[ {\left( {x + y} \right)\left( {y + z} \right)\left( {z + x} \right)} \right]^{ - 1}} \cr} $$

$$\eqalign{ & {\text{If }}\left( {{n^r} - tn + \frac{1}{4}} \right){\text{be a perfect square}} \cr & r = 2t = \pm 1 \cr & \left( {{\text{If }}t = 1} \right)\,\,{n^2} - n + \frac{1}{4} \cr & = {n^2} - 2 \times n \times \frac{1}{2} + \frac{1}{4} \cr & = {\left( {n - \frac{1}{2}} \right)^2} \cr & \left( {{\text{If }}t = - 1} \right)\,\,{n^2} + n + \frac{1}{4} \cr & = {n^2} + 2 \times n \times \frac{1}{2} + \frac{1}{4} \cr & = {\left( {n + \frac{1}{2}} \right)^2} \cr} $$

Let the numbers be x and y
According to question,
$$\eqalign{ & \left( {x > y} \right) \cr & {x^2} - {y^2} = 45 \cr & \left( {x + y} \right)\left( {x - y} \right) = 45 \cr & {\text{Make factor of 45}} \cr & {\text{15}} \times 3 \cr & \,\,9 \times 5\,\,\,\,\,\,\,\,\,\,(3\,{\text{pairs}}) \cr & 45 \times 1 \cr} $$
These pairs gives the value of x and y which satisfy the given condition.

According to question,
$$\eqalign{ & \root 3 \of {{3^n}} {\text{ = 27}} \cr & \Rightarrow {\left( {{3^n}} \right)^{\frac{1}{3}}} = {\left( 3 \right)^3} \cr & \Rightarrow {3^{\frac{n}{3}}} = {3^3} \cr & \Rightarrow \frac{n}{3} = 3 \cr & \Rightarrow n = 9 \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & \sqrt {3 + \frac{{33}}{{64}}} \div \sqrt {9 + \frac{1}{7}} \times 2\sqrt {3\frac{1}{9}} \cr & = \sqrt {\frac{{225}}{{64}}} \div \sqrt {\frac{{64}}{7}} \times 2\sqrt {\frac{{28}}{9}} \cr & = \sqrt {\frac{{225}}{{64}} \times \frac{7}{{64}} \times \frac{{28}}{9}} \times 2 \cr & = \frac{{5 \times 7}}{{8 \times 4}} \times 2 \cr & = \frac{{35}}{{16}} \cr & = 2\frac{3}{{16}} \cr} $$

$$\eqalign{ & {\text{If }}a = 3\frac{2}{3}, \cr & b = 2\frac{1}{2}, \cr & c = 4\frac{3}{4}, \cr & d = 3\frac{1}{3}, \cr & {\text{then}} \cr & {\text{Given expression ,}} \cr & \,\,\,\,\frac{{\left( {{a^2} - {b^2}} \right)}}{{\left( {{c^2} - {d^2}} \right)}} \div \frac{{\left( {a - b} \right)}}{{\left( {c - d} \right)}} \cr & = \frac{{\left( {{a^2} - {b^2}} \right)}}{{\left( {{c^2} - {d^2}} \right)}} \times \frac{{\left( {c - d} \right)}}{{\left( {a - b} \right)}} \cr & = \frac{{\left( {a + b} \right)}}{{\left( {c + d} \right)}} \cr & = \frac{{3\frac{2}{3} + 2\frac{1}{2}}}{{4\frac{3}{4} + 3\frac{1}{3}}} \cr & = \frac{{\frac{{11}}{3} + \frac{5}{2}}}{{\frac{{19}}{4} + \frac{{10}}{3}}} \cr & = \frac{{37}}{6} \times \frac{{12}}{{97}} \cr & = \frac{{74}}{{97}} \cr} $$

$$\eqalign{ & {\text{Take first part }} \cr & \frac{{4\frac{1}{7} - 2\frac{1}{4}}}{{3\frac{1}{2} + 1\frac{1}{7}}} \cr & = \frac{{\frac{{29}}{7} - \frac{9}{4}}}{{\frac{7}{2} + \frac{8}{7}}} \cr & = \frac{{\frac{{116 - 63}}{{28}}}}{{\frac{{49 + 16}}{{14}}}} \cr & = \frac{{53}}{{28}} \times \frac{{14}}{{65}} \cr & = \frac{{53}}{{130}} \cr & {\text{The second part}} \cr & \frac{1}{{2 + \frac{1}{{2 + \frac{1}{{5 - \frac{1}{5}}}}}}} \cr & = \frac{1}{{2 + \frac{1}{{2 + \frac{1}{{\frac{{25 - 1}}{5}}}}}}} \cr & = \frac{1}{{2 + \frac{1}{{2 + \frac{5}{{24}}}}}} \cr & = \frac{1}{{2 + \frac{1}{{\frac{{53}}{{24}}}}}} \cr & = \frac{1}{{2 + \frac{{24}}{{53}}}} \cr & = \frac{1}{{\frac{{106 + 24}}{{53}}}} \cr & = \frac{{53}}{{130}} \cr & {\text{According to question,}} \cr & \,\,\,\,\,\,\sqrt {\frac{{53}}{{130}} \div \frac{{53}}{{130}}} \cr & = \sqrt {\frac{{53}}{{130}} \times \frac{{130}}{{53}}} \cr & = \sqrt 1 \cr & = 1 \cr} $$

$$\eqalign{ & {\text{5P + 8C = 145785}} \cr & \Rightarrow 3(5{\text{P + 8C) = 3}} \times {\text{145785}} \cr & \Rightarrow 15{\text{P + 24C = 437355}} \cr} $$

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

$$\eqalign{ & {\text{According to question,}} \cr & {\text{3}}\frac{1}{2} - \left[ {2\frac{1}{4} \div \left\{ {1\frac{1}{4} - \frac{1}{2}\left( {1\frac{1}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right] \cr & = \frac{7}{2} - \left[ {\frac{9}{4} \div \left\{ {\frac{5}{4} - \frac{1}{2}\left( {\frac{3}{2} - \frac{1}{3} - \frac{1}{6}} \right)} \right\}} \right] \cr & = \frac{7}{2} - \left[ {\frac{9}{4} \div \left\{ {\frac{5}{4} - \frac{1}{2} \times 1} \right\}} \right] \cr & = \frac{7}{2} - \left[ {\frac{9}{4} \div \frac{3}{4}} \right] \cr & = \frac{7}{2} - \left[ {\frac{9}{4} \times \frac{4}{3}} \right] \cr & = \frac{7}{2} - 3 \cr & = \frac{{7 - 6}}{2} \cr & = \frac{1}{2} \cr} $$

$$\eqalign{ & \frac{{{x^2} + {y^2} + {z^2} - yz - zx - xy}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}{\text{ }} \cr & = \frac{{{{\left( {a + m} \right)}^2} + {{\left( {b + m} \right)}^2} + {{\left( {c + m} \right)}^2} - \left( {b + m} \right)\left( {c + m} \right) - \left( {c + m} \right)\left( {a + m} \right) - \left( {a + m} \right)\left( {b + m} \right)}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}} \cr & = \frac{{{a^2} + {m^2} + 2am + {b^2} + {m^2} + 2bm + {c^2} + {m^2} + 2cm - bc - bm - cm - {m^2} - ca - cm - am - {m^2} - ab - am - bm - {m^2}}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}} \cr & = \frac{{{a^2} + {b^2} + {c^2} - ab - bc - ca}}{{{a^2} + {b^2} + {c^2} - ab - bc - ca}} \cr & = 1 \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & \left( {\frac{{{a^3} + {b^3}}}{{{a^2} + {b^2} - ab}}} \right){\text{ = }}\left( {a + b} \right) \cr & \left[ {{\text{where}}\,a = 785,b = 345} \right] \cr & = 785 + 345 = 1220 \cr} $$

Let the highest temperature be x degrees
Then, lowest temperature
$$\eqalign{ & {\text{ = }}\left[ {\left( {1 + \frac{1}{3}} \right)\frac{x}{2}} \right]{\text{ degrees }} \cr & = \left( {\frac{4}{3} \times \frac{x}{2}} \right){\text{ degrees}} \cr & = \frac{{2x}}{3}{\text{ degrees}} \cr & \therefore x + \frac{{2x}}{3} = 100 \cr & \Leftrightarrow \frac{{5x}}{3} = 100 \cr & \Leftrightarrow x = \frac{{100 \times 3}}{5} \cr & \,\,\,\,\,\,\,\,\,\,\,\, = 60 \cr & {\text{So, lowest temperature}} \cr & {\text{ = }}\left( {\frac{2}{3} \times 60} \right){\text{degrees}} \cr & {\text{ = 40 degrees}} \cr} $$

Let the number of hats purchase be x,
$$\eqalign{ & {\text{Then, number of brown hats}} \cr & {\text{ = }}\frac{x}{4} \cr & {\text{Number of hats sold}} \cr & {\text{ = }}\frac{{2x}}{3} \cr & {\text{Number of hats left unsold}} \cr & {\text{ = }}\left( {x - \frac{{2x}}{3}} \right) = \frac{x}{3} \cr & {\text{Number of brown hats sold}} \cr & {\text{ = }}\frac{4}{5}{\text{ of }}\frac{x}{4} = \frac{x}{5} \cr & {\text{Number of brown hats left unsold}} \cr & {\text{ = }}\left( {\frac{x}{4} - \frac{x}{5}} \right) = \frac{x}{{20}} \cr & \therefore {\text{Required fraction}} \cr & {\text{ = }}\frac{{\left( {\frac{x}{{20}}} \right)}}{{\left( {\frac{x}{3}} \right)}} \cr & = \frac{x}{{20}} \times \frac{3}{x} \cr & = \frac{3}{{20}} \cr} $$

$$\eqalign{ & {\text{Given expression ,}} \cr & \frac{{{a^2} - {b^2}}}{{a + b}} = a - b \cr & = \left( {1 + \frac{1}{{1 + \frac{1}{{100}}}}} \right) - \left( {1 - \frac{1}{{1 + \frac{1}{{100}}}}} \right) \cr & = 2 \times \frac{1}{{\left( {101/100} \right)}} \cr & = 2 \times \frac{{100}}{{101}} \cr & = \frac{{200}}{{101}} \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & \,\,\,\,\,\frac{{{{\left( {75983} \right)}^2} - {{\left( {45983} \right)}^2}}}{{75983 - 45983}} \cr & = \frac{{\left( {75983 - 45983} \right)\left( {75983 + 45983} \right)}}{{(75983 - 45983)}} \cr & = \,75983 + 45983 \cr & = 121966 \cr} $$

$$\eqalign{ & {\text{According to question,}} \cr & {\text{5}}\frac{1}{3} \div 1\frac{2}{9} \times \frac{1}{4}\left( {10 + \frac{3}{{1 - \frac{1}{5}}}} \right) \cr & = \frac{{16}}{3} \div \frac{{11}}{9} \times \frac{1}{4}\left( {10 + \frac{3}{{\frac{4}{5}}}} \right) \cr & = \frac{{16}}{3} \div \frac{{11}}{9} \times \frac{1}{4}\left( {10 + \frac{{15}}{4}} \right) \cr & = \frac{{16}}{3} \times \frac{9}{{11}} \times \frac{1}{4} \times \frac{{55}}{4} \cr & = \frac{{48}}{{11}} \times \frac{1}{4} \times \frac{{55}}{4} \cr & = 15 \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & \,\,\,\,\,\,\,\,\,\,\frac{{{{\left( {\frac{7}{8} + \frac{5}{6}} \right)}^2}}}{{{{\left( {\frac{7}{8}} \right)}^2} - {{\left( {\frac{5}{6}} \right)}^2}}} \cr & = \frac{{{{\left( {\frac{7}{8} + \frac{5}{6}} \right)}^2}}}{{\left( {\frac{7}{8} + \frac{5}{6}} \right)\left( {\frac{7}{8} - \frac{5}{6}} \right)}} \cr & = \frac{{\left( {\frac{7}{8} + \frac{5}{6}} \right)}}{{\left( {\frac{7}{8} - \frac{5}{6}} \right)}} \cr & = \left( {\frac{{41}}{{24}} \times 24} \right) \cr & = 41 \cr} $$

$$\eqalign{ & {\text{Given expression,}} \cr & \frac{{{{\left( {a + b} \right)}^2} - {{\left( {a - b} \right)}^2}}}{{a b}} \cr & = \frac{{4ab}}{{ab}} \cr & = 4\,\,\,\,\,\left\{ {{\text{where }}a = 469,b = 174} \right\} \cr} $$