Simplify and write in exponential form.
(i) $ {\left( { - 2} \right)^{ - 3}} \times {\left( { - 2} \right)^{ - 4}} $ (ii) $ {p^3} \times {p^{ - 10}} $ (iii) $ {3^2} \times {3^{ - 5}} \times {3^6} $

Hint: If the bases of a power are the same in a product then the exponents must be added. The negative sign on an exponent means the reciprocal.

Complete step-by-step answer:
(i) $ {\left( { - 2} \right)^{ - 3}} \times {\left( { - 2} \right)^{ - 4}} $
Here the base is the same for both terms and the base is -2. First term has an exponent -3 and the second term has an exponent -4.
The exponents are negative so reciprocal of the terms must be taken.
 $
  {\left( { - 2} \right)^{ - 3}} = \dfrac{1}{{{{\left( { - 2} \right)}^3}}} \\
  {\left( { - 2} \right)^{ - 4}} = \dfrac{1}{{{{\left( { - 2} \right)}^4}}} \\
  $
When the bases are similar, then the exponents must be added.
 $
  {\left( { - 2} \right)^{ - 3}} \times {\left( { - 2} \right)^{ - 4}} = \dfrac{1}{{{{\left( { - 2} \right)}^3}}} \times \dfrac{1}{{{{\left( { - 2} \right)}^4}}} \\
   = \dfrac{1}{{{{\left( { - 2} \right)}^3} \times {{\left( { - 2} \right)}^4}}} \\
   = \dfrac{1}{{{{\left( { - 2} \right)}^{3 + 4}}}} \\
  \left( {\because {a^m} \times {a^n} = {a^{m + n}}} \right) \\
   = \dfrac{1}{{{{\left( { - 2} \right)}^7}}} \\
   = {\left( { - 2} \right)^{ - 7}} \\
  \left( {\because \dfrac{1}{{{a^m}}} = {a^{ - m}}} \right) \\
  $
 $ {\left( { - 2} \right)^{ - 3}} \times {\left( { - 2} \right)^{ - 4}} = {\left( { - 2} \right)^{ - 7}} $
(ii) $ {p^3} \times {p^{ - 10}} $
Here the base is the same for both terms and the base is p. First term has an exponent 3 and the second term has an exponent -10.
If the exponent is negative so reciprocal of the terms must be taken.
 $
  {\left( p \right)^{ - 10}} = \dfrac{1}{{{{\left( p \right)}^{10}}}} \\
  {p^3} \times {p^{ - 10}} = {p^3} \times \dfrac{1}{{{p^{10}}}} \\
   = \dfrac{{{p^3}}}{{{p^{10}}}} \\
   = \dfrac{1}{{{p^{10 - 3}}}} \\
  \left( {\because \dfrac{{{a^m}}}{{{a^n}}} = \dfrac{1}{{{a^{n - m}}}},n > m} \right) \\
   = \dfrac{1}{{{p^7}}} \\
   = {p^{ - 7}} \\
  \left( {\because \dfrac{1}{{{a^m}}} = {a^{ - m}}} \right) \\
  $
 $ {p^3} \times {p^{ - 10}} = {p^{ - 7}} $
(iii) $ {3^2} \times {3^{ - 5}} \times {3^6} $
Here the base is the same for all the terms and the base is 3. First term has an exponent 2, the second term has an exponent -5 and the third term has an exponent 6.
If the exponent is negative the reciprocal of the terms must be taken.
Here, the bases are similar but the exponents have different signs.
 $
  {3^{ - 5}} = \dfrac{1}{{{3^5}}} \\
  {3^2} \times {3^{ - 5}} \times {3^6} = {3^2} \times \dfrac{1}{{{3^5}}} \times {3^6} \\
   = \dfrac{{{3^2} \times {3^6}}}{{{3^5}}} \\
   = \dfrac{{{3^8}}}{{{3^5}}} \\
  \left( {\because {a^m} \times {a^n} = {a^{m + n}}} \right) \\
   = {3^{8 - 5}} \\
  \left( {\because \dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}};m > n} \right) \\
   = {3^3} \\
  $
 $ {3^2} \times {3^{ - 5}} \times {3^6} = {3^3} $

Note: A positive exponent means repeated multiplication of the base with the base, a negative exponent means repeated division of the base by the base. When the negative exponents are moved to the denominator, the exponent becomes positive.