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Question: Simplify \(\dfrac{{9 - {x^{ - 2}}}}{{3 + {x^{ - 1}}}}.\)...

Simplify 9−x−23+x−1.\dfrac{{9 - {x^{ - 2}}}}{{3 + {x^{ - 1}}}}.

Explanation

Solution

We know that for a negative exponent x−n=1xn{x^{ - n}} = \dfrac{1}{{{x^n}}}. So by using this property and the properties of algebraic rational expression we can solve this question.

Complete step by step solution:
Given
9−x−23+x−1...............................................(i)\dfrac{{9 - {x^{ - 2}}}}{{3 + {x^{ - 1}}}}...............................................\left( i \right)
So in order to solve the above given question we have to convert the numerator and denominator in such a way that we can get common terms in both the numerator and denominator and then cancel it.
For that we have to use the above mentioned property:
x−n=1xn{x^{ - n}} = \dfrac{1}{{{x^n}}}
So by using this property we have to simplify the numerator and denominator first, and then cancel the common terms if any exists.
So simplifying the numerator:
⇒9−x−2=9−1x2 ⇒9−x−2=9x2−1x2.......................(ii)  \Rightarrow 9 - {x^{ - 2}} = 9 - \dfrac{1}{{{x^2}}} \\\ \Rightarrow 9 - {x^{ - 2}} = \dfrac{{9{x^2} - 1}}{{{x^2}}}.......................\left( {ii} \right) \\\
Now simplifying the denominator:
⇒3+x−1=3+1x ⇒3+x−1=3x+1x........................(iii)  \Rightarrow 3 + {x^{ - 1}} = 3 + \dfrac{1}{x} \\\ \Rightarrow 3 + {x^{ - 1}} = \dfrac{{3x + 1}}{x}........................\left( {iii} \right) \\\
Now substituting (ii) and (iii) in (i) we get:
9−x−23+x−1=9x2−1x23x+1x =9x2−1x2×x3x+1...............(iv)  \dfrac{{9 - {x^{ - 2}}}}{{3 + {x^{ - 1}}}} = \dfrac{{\dfrac{{9{x^2} - 1}}{{{x^2}}}}}{{\dfrac{{3x + 1}}{x}}} \\\ = \dfrac{{9{x^2} - 1}}{{{x^2}}} \times \dfrac{x}{{3x + 1}}...............\left( {iv} \right) \\\
On observing (iv) we see that we can simplify 9x2−19{x^2} - 1using the property a2−b2=(a+b)(a−b):{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right):
9x2−1=(3x)2−12 =(3x+1)(3x−1).......................(v)  9{x^2} - 1 = {\left( {3x} \right)^2} - {1^2} \\\ = \left( {3x + 1} \right)\left( {3x - 1} \right).......................\left( v \right) \\\
Substituting (v) in (iv) we get:
⇒9x2−1x2×x3x+1=(3x+1)(3x−1)x2×x3x+1.....................(vi)\Rightarrow \dfrac{{9{x^2} - 1}}{{{x^2}}} \times \dfrac{x}{{3x + 1}} = \dfrac{{\left( {3x + 1} \right)\left( {3x - 1} \right)}}{{{x^2}}} \times \dfrac{x}{{3x + 1}}.....................\left( {vi} \right)
Now on observing (vi) we see that we can cancel the terms(3x+1)\left( {3x + 1} \right) and xx since it’s common to both numerator and denominator.
⇒(3x+1)(3x−1)x2×x3x+1=(3x−1)x\Rightarrow \dfrac{{\left( {3x + 1} \right)\left( {3x - 1} \right)}}{{{x^2}}} \times \dfrac{x}{{3x + 1}} = \dfrac{{\left( {3x - 1} \right)}}{x}
⇒(3x−1)x=3−1x......................(vii)\Rightarrow \dfrac{{\left( {3x - 1} \right)}}{x} = 3 - \dfrac{1}{x}......................\left( {vii} \right)
On observing (vi) we can see that it’s not possible to simply it further, therefore we can stop the process of simplification and write:
9−x−23+x−1=3−1x..................................(viii)\dfrac{{9 - {x^{ - 2}}}}{{3 + {x^{ - 1}}}} = 3 - \dfrac{1}{x}..................................\left( {viii} \right)
i.e. on simplifying 9−x−23+x−1\dfrac{{9 - {x^{ - 2}}}}{{3 + {x^{ - 1}}}}{\kern 1pt} we get 3−1x.3 - \dfrac{1}{x}.

Note: While simplifying rational expressions we should see if any general standard identities could be used and if there’s any then we should be able to convert our given expression into that form and then perform other operations which also involves the cancellation of common terms. Also the above given the method is considered suitable and easiest for similar questions containing any polynomial rational expressions.