Question

Simplify:
 $
  i){( - 4)^5} \div {( - 4)^8} \\
  ii){\left( {\dfrac{1}{{{2^3}}}} \right)^2} \\
  iii){( - 3)^4} \times {\left( {\dfrac{5}{3}} \right)^4} \\
  iv){\left( {\dfrac{2}{3}} \right)^5} \times {\left( {\dfrac{3}{4}} \right)^2} \times {\left( {\dfrac{1}{5}} \right)^2} \\
  v)({3^{ - 7}} \div {3^{10}}) \times {3^{ - 5}} \\
  vi)\dfrac{{{2^6} \times {3^2} \times {2^3} \times {3^7}}}{{{2^8} \times {3^6}}} \\
  vii){y^{a - b}} \times {y^{b - c}} \times {y^{c - a}} \\
 $

Answer 1

Hint: The above terms in the given question can be simply solved by keeping in mind that a negative number raised to the power of an even number gives a positive number always. A number to be divided can be simplified by multiplying the term with the reciprocal of the number. Also powers in subtraction can be simplified by writing the number in division with the same power.

Complete step-by-step answer:
i.Solving the first part we get that the number -4 is raised to both an odd and an even number one time. Therefore
 $
   \Rightarrow {( - 4)^5} \div {( - 4)^8} \\
   \Rightarrow \dfrac{{{{( - 4)}^5}}}{{{{( - 4)}^8}}} \\
 $
Since the numbers to be divided are same so its powers can be combined and written as:
 $ \Rightarrow {\left( { - 4} \right)^{5 - 8}} = {( - 4)^{ - 3}} $
 $
   \Rightarrow \dfrac{1}{{{{( - 4)}^3}}} = \dfrac{1}{{( - 4) \times ( - 4) \times ( - 4)}} \\
   \Rightarrow \dfrac{1}{{ - 64}} \\
 $

ii.Solving the second part by multiplying both the powers to make one power
 $
   \Rightarrow {\left( {\dfrac{1}{2}} \right)^{3 \times 2}} = {\left( {\dfrac{1}{2}} \right)^6} \\
   \Rightarrow \left( {\dfrac{1}{{2 \times 2 \times 2 \times 2 \times 2 \times 2}}} \right) = \left( {\dfrac{1}{{64}}} \right) \\
 $

iii.Solving the third part
  $ {( - 3)^4} \times {\left( {\dfrac{5}{3}} \right)^4} $
Three gets cancelled from both the numerator and denominator as they have same power and -3 also yields the same positive number since the power is a positive even number.
  $
   \Rightarrow {( - 1)^4}{(3)^4} \times {\left( {\dfrac{5}{3}} \right)^4} = {5^4} \\
  \therefore {( - 1)^4} = 1 \\
   \Rightarrow 5 \times 5 \times 5 \times 5 = 625 \; Laws of Exponents\\
 $
iv.Solving the fourth part
 $ {\left( {\dfrac{2}{3}} \right)^5} \times {\left( {\dfrac{3}{4}} \right)^2} \times {\left( {\dfrac{1}{5}} \right)^2} $
We will cancel the terms common, also we know that 4 is the square of 2 and it can be written as so:
 $
   \Rightarrow {\left( {\dfrac{2}{3}} \right)^5} \times {\left( {\dfrac{3}{{{2^2}}}} \right)^2} \times {\left( {\dfrac{1}{5}} \right)^2} \\
   \Rightarrow {\left( {\dfrac{2}{3}} \right)^5} \times \dfrac{{{{\left( 3 \right)}^2}}}{{{{(2)}^4}}} \times {\left( {\dfrac{1}{5}} \right)^2} \\
   \Rightarrow \left( {\dfrac{2}{{{3^3}}}} \right) \times 1 \times {\left( {\dfrac{1}{5}} \right)^2} \\
   \Rightarrow \dfrac{2}{{27 \times 25}} = \dfrac{2}{{675}} \\
 $

v.Solving the fifth partLaws of Exponents
 $ (\dfrac{{{3^{ - 7}}}}{{{3^{10}}}}) \times {3^{ - 5}} $
Since all the numbers are same therefore the power in division will be subtraction and the powers in multiplication will be added,
 $ {3^{ - 7 - 10 + ( - 5)}} = {3^{ - 22}} = \dfrac{1}{{{3^{22}}}} $

vi.Solving the sixth part
 $ \dfrac{{{2^6} \times {3^2} \times {2^3} \times {3^7}}}{{{2^8} \times {3^6}}} $
Let us add all the powers where bases are the same and are to be multiplied and the powers are subtracted where the similar bases are in division.
  $
   \Rightarrow \dfrac{{{2^{6 + 3}} \times {3^{2 + 7}}}}{{{2^8} \times {3^6}}} = \dfrac{{{2^9} \times {3^9}}}{{{2^8} \times {3^6}}} \\
   \Rightarrow {2^{9 - 8}} \times {3^{9 - 6}} = 2 \times {3^3} \\
   \Rightarrow 2 \times 27 = 54 \;
 $

vii..Solving the last part
 $ {y^{a - b}} \times {y^{b - c}} \times {y^{c - a}} $
Similarly dissolving the bases to the bases in division if the powers are being subtracted
 $ \Rightarrow \dfrac{{{y^a}}}{{{y^b}}} \times \dfrac{{{y^b}}}{{{y^c}}} \times \dfrac{{{y^c}}}{{{y^a}}} $
All the terms cancel each other since they are similar bases with the same power. Hence yielding;
 $ \Rightarrow 1 $

Note: Therefore all the terms are simplified above using one main concept that is adding and subtracting powers of the bases which are similar in magnitude if in multiplication and division respectively.