Decimal Fraction

Converting each of the given fractions into decimal form, we get :
$$\frac{2}{3}$$ = 0.66
$$\frac{3}{4}$$ = 0.75
$$\frac{4}{5}$$ = 0.8
$$\frac{5}{6}$$ = 0.833
Since 0.833 > 0.8 > 0.75 > 0.66
So, $$\frac{5}{6}$$ > $$\frac{4}{5}$$ > $$\frac{3}{4}$$ > $$\frac{2}{3}$$
∴ Required difference = $$\left( {\frac{5}{6} - \frac{2}{3}} \right) = \frac{1}{6}$$

$$0.121212......$$
$$ = 0.\overline {12} $$
$$ = \frac{{12}}{{99}}$$
$$ = \frac{4}{{33}}$$

Given expression :
$$\left[ {1 + \frac{1}{{1 \times 2}} + \frac{1}{{1 \times 2 \times 4}} + \frac{1}{{1 \times 2 \times 4 \times 8}} + \frac{1}{{1 \times 2 \times 4 \times 8 \times 16}}} \right]$$
$$ = \frac{{2 \times 4 \times 8 \times 16 + 4 \times 8 \times 16 + 8 \times 16 + 16 + 1}}{{2 \times 4 \times 8 \times 16}}$$
$$ = \frac{{1024 + 512 + 128 + 16 + 1}}{{1024}}$$
$$ = \frac{{1681}}{{1024}}$$
$$ = 1.6416$$

Given expression :
$$\frac{{{{\left( {0.013} \right)}^3} + 0.000000343}}{{{{\left( {0.013} \right)}^2} - 0.000091 + 0.000049}}$$
$$a = 0.013 \text{ and } b = 0.007 $$
$$ = \left( {\frac{{{a^3} + {b^3}}}{{{a^2} - ab + {b^2}}}} \right)$$
$$ = a + b$$
$$ = 0.013 + 0.007$$
$$ = 0.020$$

Given expression :
$$\left[ {\frac{{8{{\left( {3.75} \right)}^3} + 1}}{{{{\left( {7.5} \right)}^2} - 6.5}}} \right]$$
$$ = \frac{{{{\left( {2 \times 3.75} \right)}^3} + {1^3}}}{{{{\left( {7.5} \right)}^2} - \left( {7.5 \times 1} \right) + {1^2}}}$$
$$ = \frac{{{{\left( {7.5} \right)}^3} + {1^3}}}{{{{\left( {7.5} \right)}^2} - \left( {7.5 \times 1} \right) + {1^2}}}$$
$$ = \left( {\frac{{{a^3} + {b^3}}}{{{a^2} - ab + {b^2}}}} \right) $$
$$ = \frac{{\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)}}{{\left( {{a^2} - ab + {b^2}} \right)}} $$
$$ = \left( {a + b} \right)$$
$$ = \left( {7.5 + 1} \right)$$
$$ = 8.5$$

The given expression can be written in this form also
N = $${\text{ 0}}{\text{.39}}\overline {{\text{39}}} $$ ..... (i)
Multiply equation (i) with 100 on both sides.
100N = $${\text{39}}{\text{.}}\overline {{\text{39}}} $$ ..... (ii)
Subtracting equation (i) from (ii) we get,
⇒ 100N - N = $${\text{39}}\overline {{\text{.39}}} $$  - $${\text{0}}\overline {{\text{.39}}} $$
⇒ 99N = 39
⇒ N = $$\frac{39}{99}$$ = $$\frac{13}{33}$$

Given expression :
$$ = \frac{{{{\left( {0.051} \right)}^3} + {{\left( {0.041} \right)}^3}}}{{{{\left( {0.051} \right)}^2} - \left( {0.051 \times 0.041} \right) + {{\left( {0.041} \right)}^2}}}$$
Let 0.051 = $$a$$ and 0.041 = $$b$$
$$ = \left( {\frac{{{a^3} + {b^3}}}{{{a^2} - ab + {b^2}}}} \right)$$
$$ = (a + b)$$
$$ = \left( {0.051 + 0.041} \right)$$
$$ = 0.092$$

Given expression :
$$ = \frac{{{{\left( {10.3} \right)}^3} + {1^3}}}{{{{\left( {10.3} \right)}^2} - \left( {10.3 \times 1} \right) + {1^2}}}$$
$$ = \left( {\frac{{{a^3} + {b^3}}}{{{a^2} - ab + {b^2}}}} \right)$$
$$ = (a + b)$$
$$ = \left( {10.3 + 1} \right)$$
$$ = 11.3$$

Given expression :
$$ = \frac{{{{\left( {4.53 - 3.07} \right)}^3} + {{\left( {3.07 - 2.15} \right)}^3} + {{\left( {2.15 - 4.53} \right)}^3}}}{{\left( {4.53 - 3.07} \right)\left( {3.07 - 2.15} \right)\left( {2.15 - 4.53} \right)}}$$
Let (4.53 - 3.07) = $$a$$, (3.07 - 2.15) = $$b$$ and (2.15 - 4.53) = $$c$$
$$ = \frac{{{a^3} + {b^3} + {c^3}}}{{abc}}$$
$$ = \frac{{3abc}}{{abc}}$$
$$ = 3$$
[∵ If a + b + c = 0, a ³ + b³ + c³ = 3abc]