Question

Simplify: $\left[ {\dfrac{{x + 2}}{{x - 2}} - \dfrac{{x - 2}}{{x + 2}} - \dfrac{{8x}}{{{x^2} - 16}}} \right] \div \dfrac{{8x}}{{{x^4} - 16}}$.

Answer 1

Hint:
 We can simplify the dividend and divisor separately. To simplify the dividend, we can take the ${1}^{\text{st}}$ 2 terms and simplify it by finding the LCM and using the identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$. Then we can simplify the remaining terms by taking the LCM and doing further calculation. Then we can take the divisor and expand its denominator using the identity $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ . Then we can write them as fractions and cancel out the common terms to get the simplified expression.

Complete step by step solution:
We can take the term inside the bracket,
Let,
We can take the LCM of the 1st 2 terms,
 $ \Rightarrow A = \left[ {\dfrac{{{{\left( {x + 2} \right)}^2} - {{\left( {x - 2} \right)}^2}}}{{\left( {x + 2} \right)\left( {x - 2} \right)}} - \dfrac{{8x}}{{{x^2} - 16}}} \right]$
We know that, $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ . Using this we can simplify the numerator and denominator,
 $ \Rightarrow A = \left[ {\dfrac{{\left( {x + 2 + x - 2} \right)\left( {x + 2 - x + 2} \right)}}{{{x^2} - {2^2}}} - \dfrac{{8x}}{{{x^2} - 16}}} \right]$
On simplification, we get,
 $ \Rightarrow A = \left[ {\dfrac{{\left( {2x} \right)\left( 4 \right)}}{{{x^2} - 4}} - \dfrac{{8x}}{{{x^2} - 16}}} \right]$
 $ \Rightarrow A = \left[ {\dfrac{{8x}}{{{x^2} - 4}} - \dfrac{{8x}}{{{x^2} - 16}}} \right]$
We can take the term 8x outside the bracket.
 $ \Rightarrow A = 8x\left[ {\dfrac{1}{{{x^2} - 4}} - \dfrac{1}{{{x^2} - 16}}} \right]$
Now we can take the LCM,
 \[ \Rightarrow A = 8x\left[ {\dfrac{{{x^2} - 16 - \left( {{x^2} - 4} \right)}}{{\left( {{x^2} - 4} \right)\left( {{x^2} - 16} \right)}}} \right]\]
On further simplification, we get,
 \[ \Rightarrow A = 8x\left[ {\dfrac{{{x^2} - 16 - {x^2} + 4}}{{\left( {{x^2} - 4} \right)\left( {{x^2} - 16} \right)}}} \right]\]
 \[ \Rightarrow A = 8x\left[ {\dfrac{{ - 12}}{{\left( {{x^2} - 4} \right)\left( {{x^2} - 16} \right)}}} \right]\]
Hence we have,
 \[ \Rightarrow \left[ {\dfrac{{x + 2}}{{x - 2}} - \dfrac{{x - 2}}{{x + 2}} - \dfrac{{8x}}{{{x^2} - 16}}} \right] = 8x\left[ {\dfrac{{ - 12}}{{\left( {{x^2} - 4} \right)\left( {{x^2} - 16} \right)}}} \right]\]… (1)
Now we can take the divisor,
 $B = \dfrac{{8x}}{{{x^4} - 16}}$
We know that, $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ . Using this we can simplify the denominator,
 $ \Rightarrow B = \dfrac{{8x}}{{\left( {{x^2} - 4} \right)\left( {{x^2} + 4} \right)}}$
Hence we have,
 $ \Rightarrow \dfrac{{8x}}{{{x^4} - 16}} = \dfrac{{8x}}{{\left( {{x^2} - 4} \right)\left( {{x^2} + 4} \right)}}$ … (2)
Now the expression,
  $I = \left[ {\dfrac{{x + 2}}{{x - 2}} - \dfrac{{x - 2}}{{x + 2}} - \dfrac{{8x}}{{{x^2} - 16}}} \right] \div \dfrac{{8x}}{{{x^4} - 16}}$
On substituting (1) and (2), we get,
 $ \Rightarrow I = \dfrac{{8x\dfrac{{ - 12}}{{\left( {{x^2} - 4} \right)\left( {{x^2} - 16} \right)}}}}{{\dfrac{{8x}}{{\left( {{x^2} - 4} \right)\left( {{x^2} + 4} \right)}}}}$
We can cancel all the common term, then we will get,
 $ \Rightarrow I = \dfrac{{\dfrac{{ - 12}}{{\left( {{x^2} - 16} \right)}}}}{{\dfrac{1}{{\left( {{x^2} + 4} \right)}}}}$
On rearranging, we get,
 $ \Rightarrow I = \dfrac{{ - 12\left( {{x^2} + 4} \right)}}{{\left( {{x^2} - 16} \right)}}$

Therefore, the required simplified expression is $\dfrac{{ - 12\left( {{x^2} + 4} \right)}}{{\left( {{x^2} - 16} \right)}}$

Note:
We used the concepts of algebra to solve the problem. The important identity we used is $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$ . While simplifying the numerator, we must take only 2 terms at a time. Otherwise it will become complicated and we may go wrong. We must take care of the power of x before cancelling. We must verify that the sign between the variable and constant is negative before it is factored.