Question

How do you simplify \[\dfrac{{{\left( {{x}^{-3}} \right)}^{4}}{{x}^{4}}}{2{{x}^{-3}}}\] and write it using only positive exponents?

Answer 1

Hint: We use cancellation of similar terms to simplify \[\dfrac{{{\left( {{x}^{-3}} \right)}^{4}}{{x}^{4}}}{2{{x}^{-3}}}.\] In an algebraic fraction, like arithmetic fraction, we consider the power of the variables in both the numerator and the denominator. After cutting off the similar terms up to the same power from the numerator and the denominator, the rest of the terms will remain as they are. Remember \[{{x}^{-n}}=\dfrac{1}{{{x}^{n}}}.\] Apply this to make the fraction only of positive exponents. Also remember \[{{\left( \dfrac{1}{x} \right)}^{n}}=\dfrac{1}{{{x}^{n}}}.\] Furthermore, \[{{\left( \dfrac{1}{{{x}^{n}}} \right)}^{m}}=\dfrac{1}{{{\left( {{x}^{n}} \right)}^{m}}}=\dfrac{1}{{{x}^{nm}}}.\] Recall that we have learnt \[{{x}^{n}}.{{x}^{m}}={{x}^{n+m}}.\] \[\]

Complete step by step solution:
Consider the algebraic fraction \[\dfrac{{{\left( {{x}^{-3}} \right)}^{4}}{{x}^{4}}}{2{{x}^{-3}}}.\]
We have to cut off the similar terms from the numerator and the denominator.
We have \[{{x}^{-3}}\] in both the numerator and the denominator.
So, \[\dfrac{{{({{x}^{-3}})}^{4}}{{x}^{4}}}{2{{x}^{-3}}}=\dfrac{{{x}^{-3}}.{{x}^{-3}}.{{x}^{-3}}.{{x}^{-3}}.{{x}^{4}}}{2{{x}^{-3}}}\]
                         \[=\dfrac{{{x}^{-3}}.{{x}^{-3}}.{{x}^{-3}}.{{x}^{4}}}{2},\] after cancelling \[{{x}^{-3}}\] from both the numerator and the denominator.
Now we get the equation as, \[\dfrac{{{({{x}^{-3}})}^{4}}{{x}^{4}}}{2{{x}^{-3}}}=\dfrac{{{({{x}^{-3}})}^{3}}{{x}^{4}}}{2}\]
We apply the identity, ${{\left( {{x}^{-m}} \right)}^{n}}={{x}^{-mn}}$
So, this can be written as \[\dfrac{{{({{x}^{-3}})}^{4}}{{x}^{4}}}{2{{x}^{-3}}}=\dfrac{({{x}^{-9}}){{x}^{4}}}{2}\]
Again, we use another identity, ${{x}^{-m}}{{x}^{n}}={{x}^{-m+n}}$
Now we get,
$\Rightarrow \dfrac{{{\left( {{x}^{-3}} \right)}^{4}}{{x}^{4}}}{2{{x}^{-3}}}=\dfrac{{{x}^{-9+4}}}{2}=\dfrac{{{x}^{-5}}}{2}$
The simplified form of the given algebraic fraction \[\dfrac{{{({{x}^{-3}})}^{4}}{{x}^{4}}}{2{{x}^{-3}}}=\dfrac{{{x}^{-5}}}{2}.\]
Now we have to write it using only positive exponents.
We can write, \[\dfrac{{{({{x}^{-3}})}^{4}}{{x}^{4}}}{2{{x}^{-3}}}=\dfrac{{{x}^{-5}}}{2}=\dfrac{\left( \dfrac{1}{{{x}^{5}}} \right)}{2}=\dfrac{1}{2}\dfrac{1}{{{x}^{5}}}=\dfrac{1}{2{{x}^{5}}}\]

Hence, we get \[\dfrac{{{({{x}^{-3}})}^{4}}{{x}^{4}}}{2{{x}^{-3}}}=\dfrac{1}{2{{x}^{5}}}.\]

Note: We cancel the terms only if it's multiplied with \[1\] or any other terms in both the numerator and the denominator. We are not allowed to do the cancellation of terms if the terms in the numerator or denominator are added to or subtracted from other terms (same variable or different variables) in the numerator or the denominator.
This can also be done in another way:
\[\dfrac{{{\left( {{x}^{-3}} \right)}^{4}}{{x}^{4}}}{2{{x}^{-3}}}=\dfrac{{{\left( \dfrac{1}{{{x}^{3}}} \right)}^{4}}{{x}^{4}}}{2\dfrac{1}{{{x}^{3}}}}=\dfrac{\dfrac{1}{{{\left( {{x}^{3}} \right)}^{4}}}{{x}^{4}}}{2\dfrac{1}{{{x}^{3}}}}=\dfrac{\dfrac{1}{{{x}^{12}}}{{x}^{4}}}{2\dfrac{1}{{{x}^{3}}}}\]
           \[=\dfrac{\dfrac{1}{{{x}^{12-4}}}}{2\dfrac{1}{{{x}^{3}}}},\] \[\left[ \dfrac{1}{{{x}^{n}}}{{x}^{m}}=\dfrac{1}{{{x}^{n-m}}} \right]\] \[\]
          \[=\dfrac{\dfrac{1}{{{x}^{8}}}}{2\dfrac{1}{{{x}^{3}}}}=\dfrac{{{\left( \dfrac{1}{x} \right)}^{8}}}{2{{\left( \dfrac{1}{x} \right)}^{3}}}=\dfrac{\dfrac{1}{x}\dfrac{1}{x}\dfrac{1}{x}\dfrac{1}{x}\dfrac{1}{x}\dfrac{1}{x}\dfrac{1}{x}\dfrac{1}{x}}{2\dfrac{1}{x}\dfrac{1}{x}\dfrac{1}{x}}\]
Cancelling \[\dfrac{1}{x}\dfrac{1}{x}\dfrac{1}{x}\],
      \[=\dfrac{\dfrac{1}{x}\dfrac{1}{x}\dfrac{1}{x}\dfrac{1}{x}\dfrac{1}{x}}{2}=\dfrac{{{\left( \dfrac{1}{x} \right)}^{5}}}{2}=\dfrac{\dfrac{1}{{{x}^{5}}}}{2}=\dfrac{1}{2}\dfrac{1}{{{x}^{5}}}=\dfrac{1}{2{{x}^{5}}}.\]