Question

If $ a - \frac{1}{a} = 8 $ and $ a \ne 0 $ ; find $ {a^2} - \frac{1}{{{a^2}}} $ ...

Answer 1

Hint: To simplify this question , we need to solve it step by step . Here we are going to expand the Left hand side part and perform some calculations to simplify the given equation by somewhere using equivalent equations and algebraic identities . Equivalent equations are said to be algebraic equations that may have the same solutions if we add or subtract the same number to both sides of an equation - Left hand side or Right hand side of the equal to sign . Or we can multiply or divide the same number to both sides of an equation - Left hand side or Right hand side of the equal to sign with the method of simplification .

Complete step-by-step answer:
The question given to us is $ a - \frac{1}{a} = 8 $ and $ a \ne 0 $ and we have to find $ {a^2} - \frac{1}{{{a^2}}} $ .
We will first perform squaring both the sides and we will get ,
 $ {\left( {a - \frac{1}{a}} \right)^2} = {8^2} $
Now , as we can see that an algebraic identity can be applied and solved further that is $ {(A - B)^2} = {A^2} - 2AB + {B^2} $
We will consider A = a and $ B = \frac{1}{a} $ , substituting these in the given identity we will get –
\[
  {\left( {a - \frac{1}{a}} \right)^2} = {8^2} \\
  {a^2} - 2a\left( {\frac{1}{a}} \right) + {\left( {\frac{1}{a}} \right)^2} = 64 \\
\]
Simplifying the equation by performing simple calculations like canceling out the ‘a’ .
\[{a^2} - 2 + {\left( {\frac{1}{a}} \right)^2} = 64\]
Rewriting the equation in the correct format , we get –
\[{a^2} + {\left( {\frac{1}{a}} \right)^2} = 64 + 2\]
\[{a^2} + {\left( {\frac{1}{a}} \right)^2} = 66\]--------------------- equation 1
In order to get the value of $ {a^2} - \frac{1}{{{a^2}}} $ , we should remember the algebraic identity for the same which is as follows –
  $ {A^2} - {B^2} = (A + B)(A - B) $
 $ {a^2} - \frac{1}{{{a^2}}} = \left( {a + \frac{1}{a}} \right)\left( {a - \frac{1}{a}} \right) $ -------------- equation 2 .
We are given the value of $ a - \frac{1}{a} = 8 $ , we need to only find the value of $ \left( {a + \frac{1}{a}} \right) $ .
Similarly , we will calculate for $ {\left( {a + \frac{1}{a}} \right)^2} $ by expanding and applying the algebraic identity as follows –
 $ {(A + B)^2} = {A^2} + 2AB + {B^2} $
 $ {\left( {a + \frac{1}{a}} \right)^2} = {\left( a \right)^2} + {\left( {\frac{1}{a}} \right)^2} + 2\left( a \right)\left( {\frac{1}{a}} \right) $
 $ {\left( {a + \frac{1}{a}} \right)^2} = {\left( a \right)^2} + {\left( {\frac{1}{a}} \right)^2} + 2 $ ------equation 3
Now as we already have the value of \[{a^2} + {\left( {\frac{1}{a}} \right)^2} = 66\] as solved above named as equation 1 substitute the value as it is in equation 3.
 $ {\left( {a + \frac{1}{a}} \right)^2} = 66 + 2 = 68 $
But we want the value of $ \left( {a + \frac{1}{a}} \right) $ , so we will take square root and we will get ,
 $ \left( {a + \frac{1}{a}} \right) = \sqrt {68} = \pm 2\sqrt {17} $ .
Now substitute the value as it is in equation 2 ,
 $ {a^2} - \frac{1}{{{a^2}}} = \left( {a + \frac{1}{a}} \right)\left( {a - \frac{1}{a}} \right) $
 $ {a^2} - \frac{1}{{{a^2}}} = \left( { \pm 2\sqrt {17} } \right)\left( 8 \right) = \pm 16\sqrt {17} $
So, the correct answer is “ $ \pm 16\sqrt {17} $ ”.

Note: In equivalent equations which have identical solutions we can perform multiplication or division by the same non-zero number both L.H.S. and R.H.S. of an equation
In an equivalent equation which has an identical solution we can raise the same odd power to both L.H.S. and R.H.S. of an equation .
Cross check the answer and always keep the final answer simplified .
Remember the algebraic identities and apply appropriately .