Surds And Indices

$${\text{Given}}\,{\text{Expression}}$$
$$ = \frac{{{{\left( {243} \right)}^{n/5}} \times {3^{2n + 1}}}}{{{9^n} \times {3^{n - 1}}}}$$
$$ = \frac{{{{\left( {{3^5}} \right)}^{n/5}} \times {3^{2n + 1}}}}{{{{\left( {{3^2}} \right)}^n} \times {3^{n - 1}}}}$$
$$ = \frac{{\left( {{3^{5 \times \left( {n/5} \right)}} \times {3^{2n + 1}}} \right)}}{{\left( {{3^{2n}} \times {3^{n - 1}}} \right)}}$$
$$ = \frac{{{3^n} \times {3^{2n + 1}}}}{{{3^{2n}} \times {3^{n - 1}}}}$$
$$ = \frac{{{3^{\left( {n + 2n + 1} \right)}}}}{{{3^{\left( {2n + n - 1} \right)}}}}$$
$$ = \frac{{{3^{3n + 1}}}}{{{3^{3n - 1}}}}$$
$$ = {3^{\left( {3n + 1 - 3n + 1} \right)}}$$
$$ = {3^2}$$
$$ = 9$$

$$\frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}}$$
$$ = \frac{1}{{\left( {1 + \frac{{{a^n}}}{{{a^m}}}} \right)}} + \frac{1}{{\left( {1 + \frac{{{a^m}}}{{{a^n}}}} \right)}}$$
$$ = \frac{{{a^m}}}{{\left( {{a^m} + {a^n}} \right)}} + \frac{{{a^n}}}{{\left( {{a^m} + {a^n}} \right)}}$$
$$ = \frac{{\left( {{a^m} + {a^n}} \right)}}{{\left( {{a^m} + {a^n}} \right)}}$$
$$ = 1$$

We know that 11² = 121.
Putting m = 11 and n = 2, we get:
(m - 1)^{n + 1} = (11 - 1)^{(2 + 1)} = 10³ = 1000

$${\text{Given Exp}}{\text{.}}$$
$$ = {x^{\left( {b - c} \right)\left( {b + c - a} \right)}}.{x^{\left( {c - a} \right)\left( {c + a - b} \right)}}.{x^{\left( {a - b} \right)\left( {a + b - c} \right)}}$$
  $$ = {x^{\left( {b - c} \right)\left( {b + c} \right) - a\left( {b - c} \right)}}.$$    $${x^{\left( {c - a} \right)\left( {c + a} \right) - b\left( {c - a} \right)}}.$$   $${x^{\left( {a - b} \right)\left( {a + b} \right) - c\left( {a - b} \right)}}$$
$$ = {x^{\left( {{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}} \right)}}.{x^{ - a\left( {b - c} \right) - b\left( {c - a} \right) - c\left( {a - b} \right)}}$$
$$ = {{x^0} . {x^\left( -ab + ac - bc + ba - ca + cb \right)}} $$
$$ = \left( {{x^0} \times {x^0}} \right)$$
$$ = \left( {1 \times 1} \right)$$
$$ = 1$$

$${\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2}$$
$$ = x + \frac{1}{x} - 2$$
$$ = \left( {3 + 2\sqrt 2 } \right) + \frac{1}{{\left( {3 + 2\sqrt 2 } \right)}} - 2$$
$$ = \left( {3 + 2\sqrt 2 } \right) + \frac{1}{{\left( {3 + 2\sqrt 2 } \right)}} \times \frac{{\left( {3 - 2\sqrt 2 } \right)}}{{\left( {3 - 2\sqrt 2 } \right)}} - 2$$
$$ = \left( {3 + 2\sqrt 2 } \right) + \left( {3 - 2\sqrt 2 } \right) - 2$$
$$ = 4$$
$$\therefore \left( {\sqrt x - \frac{1}{{\sqrt x }}} \right) = 2$$

3⁸ × 3⁴
= 3^{(8 + 4)}
= 3¹²
= (3⁶)²
= (729)²

$$\frac{{343 \times 49}}{{216 \times 16 \times 81}}$$
$$ = \frac{{{7^3} \times {7^2}}}{{{6^3} \times {2^4} \times {3^4}}}$$
$$ = \frac{{{7^{\left( {3 + 2} \right)}}}}{{{6^3} \times {{\left( {2 \times 3} \right)}^4}}}$$
$$ = \frac{{{7^5}}}{{{6^3} \times {6^4}}}$$
$$ = \frac{{{7^5}}}{{{6^{\left( {3 + 4} \right)}}}}$$
$$ = \frac{{{7^5}}}{{{6^7}}}$$

$$\frac{{16 \times 32}}{{9 \times 27 \times 81}}$$
$$ = \frac{{{2^4} \times {2^5}}}{{{3^2} \times {3^3} \times {3^4}}}$$
$$ = \frac{{{2^{\left( {4 + 5} \right)}}}}{{{3^{\left( {2 + 3 + 4} \right)}}}}$$
$$ = \frac{{{2^9}}}{{{3^9}}}$$
$$ = {\left( {\frac{2}{3}} \right)^9}$$

$$\sqrt 2 = 1.414$$
$$ ⇒ \sqrt 8 {\text{ + 2}}\sqrt {32} - 3\sqrt {128} {\text{ + 4}}\sqrt {50} $$
$$ ⇒ 2\sqrt 2 + 2 \times 4\sqrt 2 - 3 \times 8\sqrt 2 + 4 \times 5\sqrt 2 $$
$$ ⇒ 2\sqrt 2 + 8\sqrt 2 - 24\sqrt 2 + 20\sqrt 2 $$
$$ ⇒ 6\sqrt 2 $$
$$ ⇒ 6 \times 1.414$$
$$ ⇒ 8.484{\text{ }}$$

$${\text{Let}}\,{9^3} \times {\left( {81} \right)^2} \div {\left( {27} \right)^3} = {\left( 3 \right)^x}{\text{then}}$$
$$ ⇒ {\left( 3 \right)^x}{\text{ = }}\frac{{{{\left( {{3^2}} \right)}^3} \times {{\left( {{3^4}} \right)}^2}}}{{{{\left( {{3^3}} \right)}^3}}}$$
$$ ⇒ {\left( 3 \right)^x} = \frac{{{3^{\left( {2 \times 3} \right)}} \times {3^{\left( {4 \times 2} \right)}}}}{{{3^{\left( {3 \times 3} \right)}}}}$$
$$ ⇒ {\left( 3 \right)^x} = \frac{{{3^6} \times {3^8}}}{{{3^9}}}$$
$$ ⇒ {\left( 3 \right)^x} = \frac{{{3^{\left( {6 + 8} \right)}}}}{{{3^9}}}$$
$$ ⇒ {\left( 3 \right)^x} = \frac{{{3^{14}}}}{{{3^9}}}$$
$$ ⇒ {\left( 3 \right)^x} = {3^{\left( {14 - 9} \right)}}$$
$$ ⇒ {\left( 3 \right)^x} = {3^5}$$
$$ ⇒ {\left( 3 \right)^x} = 5$$