Solution:
$$\eqalign{ {a^2} + {b^2} + {c^2} + 27 = 6\left( {a + b + c} \right) \cr {\left( {a - 3} \right)^2} + {\left( {b - 3} \right)^2} + {\left( {c - 3} \right)^2} = 0 \cr a = 3,\,b = 3,\,c = 3 \cr \root 3 \of {{a^3} + {b^3} - {c^3}} \cr = \root 3 \of {27 + 27 - 27} \cr = \root 3 \of {27} \cr = 3 \cr} $$

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

Solution:
$$\eqalign{ {\text{According to the question,}} \cr {\text{If ,}}2s = a + b + c \cr \Leftrightarrow s = \frac{{a + b + c}}{2} \cr {\text{Let, }} \cr a = 10 \cr b = 10 \cr c = 10 \cr \because s = \frac{{a + b + c}}{2} \cr \Rightarrow s = \frac{{10 + 10 + 10}}{2} \cr \Rightarrow s = \frac{{30}}{2} \cr \Rightarrow s = 15 \cr \therefore s\left( {s - c} \right) + \left( {s - a} \right)\left( {s - b} \right) \cr = 15\left( {15 - 10} \right) + \left( {15 - 10} \right)\left( {15 - 10} \right) \cr = 75 + 25 \cr = 100 \cr {\text{Now check from option, }} \cr {\text{Option 'A' }}ab = 10 \times 10 \cr \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 100\left( {{\text{Satisfied}}} \right) \cr} $$

Solution:
$$\eqalign{ \frac{{2x - y}}{{x + 2y}}\, \times \frac{1}{2}\left( {{\text{Cross multiply}}} \right) \cr \Rightarrow 4x - 2y = x + 2y \cr \Rightarrow 3x = 4y \cr \Rightarrow x:y = 4:3 \cr \therefore \frac{{3x - y}}{{3x + y}}{\text{ }} \cr {\text{ = }}\frac{{3 \times 4 - 3}}{{3 \times 4 + 3}} \cr {\text{ = }}\frac{{12 - 3}}{{12 + 3}} \cr {\text{ = }}\frac{9}{{15}}{\text{ }} \cr {\text{ = }}\frac{3}{5} \cr} $$

Solution:
1 + 9r² + 81r⁴ = 256
1 + 3r + 9r² = 32
So, 1 - 3r + 9r² $$ = \frac{{256}}{{32}} = 8$$

Solution:
$$\eqalign{ {\text{According to the question,}} \cr \Rightarrow {\left( {x + 3} \right)^3} + \frac{1}{{{{\left( {x + 3} \right)}^3}}} \cr {\text{Let, }}t = x + 3,{\text{ }}x = t - 3 \cr \Rightarrow {x^2} + x = 5{\text{, }}\left( {{\text{Given}}} \right) \cr \Rightarrow \left( {x + 1} \right)x = 5 \cr \Rightarrow \left( {t - 3 + 1} \right).\left( {t - 3} \right) = 5 \cr \Rightarrow \left( {t - 2} \right).\left( {t - 3} \right) = 5 \cr \Rightarrow {t^2} - 3t - 2t + 6 = 5 \cr \Rightarrow {t^2} - 5t = - 1 \cr \Rightarrow t + \frac{1}{t} = 5 \cr \Rightarrow {\left( {x + 3} \right)^3} + \frac{1}{{{{\left( {x + 3} \right)}^3}}} \cr \Rightarrow {t^3} + \frac{1}{{{t^3}}} \cr \Rightarrow \left( {t + \frac{1}{t}} \right)^3 - 3 \times t \times \frac{1}{t}\left( {t + \frac{1}{t}} \right) \cr \Rightarrow {5^3} - 3 \times 5 \cr \Rightarrow 125 - 15 \cr \Rightarrow 110 \cr \therefore {\left( {x + 3} \right)^3} + \frac{1}{{{{\left( {x + 3} \right)}^3}}} = 110 \cr} $$

Solution:
$$a - \frac{1}{{a - 5}} = 10$$
We can write it is as (subtracting 5 from both sides)
$$\eqalign{ a - 5 - \frac{1}{{a - 5}} = 10 - 5 \cr {\text{Now, }}\left( {a - 5} \right) - \frac{1}{{\left( {a - 5} \right)}} = 5 \cr} $$
So, take the cube of this equation,
$$\eqalign{ \left[ {{{\left( {a - 5} \right)}^3} - \frac{1}{{{{\left( {a - 5} \right)}^3}}}} \right] = 125 + 3 \times 5 \cr \left[ {{{\left( {a - 5} \right)}^3} - \frac{1}{{{{\left( {a - 5} \right)}^3}}}} \right] = 140 \cr} $$