Question

Find the value of the following expression:
 ${{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}$

Answer 1

Hint:In this problem we can use the following formula:
 ${{x}^{2}}+{{y}^{2}}={{\left( x+y \right)}^{2}}-2{{x}^{2}}{{y}^{2}}$ . Or we can use,
${{x}^{2}}+{{y}^{2}}=\dfrac{1}{2}\left[ {{\left( x+y \right)}^{2}}+{{\left( x-y \right)}^{2}} \right]$
We use substitution of the terms in the bracket with some variables to make simplification easier then finally substitute the original terms in the answer.

Complete Step-by-step answer:
Let us first take the given expression:
${{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}$
We can write it as,
${{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}={{\left( {{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{2}} \right)}^{2}}+{{\left( {{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{2}} \right)}^{2}}$
Let us assign ${{a}^{2}}=p$ and $\sqrt{{{a}^{2}}-1}=q$ to simplify the expression. So our expression will become:
${{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}={{\left( {{\left( p+q \right)}^{2}} \right)}^{2}}+{{\left( {{\left( p-q \right)}^{2}} \right)}^{2}}$
Now let us apply the formula, ${{x}^{2}}+{{y}^{2}}={{\left( x+y \right)}^{2}}-2{x}{{y}}$. By applying the formula, we will get:
${{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}={{\left( {{\left( p+q \right)}^{2}}+{{\left( p-q \right)}^{2}} \right)}^{2}}-2{{\left( p+q \right)}^{2}}{{\left( p-q \right)}^{2}}$
Now, we can again apply the formula ${{x}^{2}}+{{y}^{2}}={{\left( x+y \right)}^{2}}-2{{x}^{2}}{{y}^{2}}$ in the part $\left( {{\left( p+q \right)}^{2}}+{{\left( p-q \right)}^{2}} \right)$, we get
${{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}={{\left( {{\left( p+q+p-q \right)}^{2}}-2\left( p+q \right)\left( p-q \right) \right)}^{2}}-2{{\left( \left( p+q \right)\left( p-q \right) \right)}^{2}}$ ,
By applying the formula ${{p}^{2}}-{{q}^{2}}=\left( p+q \right)\left( p-q \right)$ we will get,
$\begin{align}
  & {{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}={{\left( {{\left( 2p \right)}^{2}}-2\left( {{p}^{2}}-{{q}^{2}} \right) \right)}^{2}}-2{{\left( {{p}^{2}}-{{q}^{2}} \right)}^{2}} \\
 & \Rightarrow {{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}={{\left( 4{{p}^{2}}-2{{p}^{2}}+2{{q}^{2}} \right)}^{2}}-2{{\left( {{p}^{2}}-{{q}^{2}} \right)}^{2}} \\
\end{align}$
By applying the formula ${{\left( x-y \right)}^{2}}={{x}^{2}}-2xy+{{y}^{2}}$ we will get,
\[\begin{align}
  & {{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}={{\left( 2{{p}^{2}}+2{{q}^{2}} \right)}^{2}}-2\left( {{\left( {{p}^{2}} \right)}^{2}}-2{{p}^{2}}{{q}^{2}}+{{\left( {{q}^{2}} \right)}^{2}} \right) \\
 & \Rightarrow {{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}={{\left( 2{{p}^{2}}+2{{q}^{2}} \right)}^{2}}-2\left( {{p}^{4}}-2{{p}^{2}}{{q}^{2}}+{{q}^{4}} \right) \\
\end{align}\]
Now we will apply the formula, ${{\left( x+y \right)}^{2}}={{x}^{2}}+2xy+{{y}^{2}}$, the above expression can be written as
$\begin{align}
  & {{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}=\left( {{\left( 2{{p}^{2}} \right)}^{2}}+2\times \left( 2{{p}^{2}} \right)\times \left( 2{{q}^{2}} \right)+{{\left( 2{{q}^{2}} \right)}^{2}} \right)-2{{p}^{4}}+4{{p}^{2}}{{q}^{2}}-2{{q}^{4}} \\
 & \Rightarrow {{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}=4{{p}^{4}}+8{{p}^{2}}{{q}^{2}}+4{{q}^{4}}-2{{p}^{4}}+4{{p}^{2}}{{q}^{2}}-2{{q}^{4}} \\
\end{align}$
By grouping the terms with same variables together and adding or subtracting them we will get,
$\begin{align}
  & {{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}=\left( 4{{p}^{4}}-2{{p}^{4}} \right)+\left( 8{{p}^{2}}{{q}^{2}}+4{{p}^{2}}{{q}^{2}} \right)+\left( 4{{q}^{4}}-2{{q}^{4}} \right) \\
 & \Rightarrow {{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}=2{{p}^{4}}+12{{p}^{2}}{{q}^{2}}+2{{q}^{4}} \\
\end{align}$
By substituting the value of $p$ and $q$we get,
${{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}=2{{\left( {{a}^{2}} \right)}^{4}}+12{{\left( {{a}^{2}} \right)}^{2}}{{\left( \sqrt{{{a}^{2}}-1} \right)}^{2}}+2{{\left( \sqrt{{{a}^{2}}-1} \right)}^{4}}$
Now, apply the Laws of Exponents to simplify the above expression. Mainly the Power of a Power Rule, which is ${{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}}$ and multiplication Rule, which is ${{a}^{x}}\times {{a}^{y}}={{a}^{x+y}}$, the above expression can be written as,
$\begin{align}
  & {{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}=2{{a}^{2\times 4}}+12{{a}^{2\times 2}}{{\left( {{a}^{2}}-1 \right)}^{\dfrac{1}{2}\times 2}}+2{{\left( {{a}^{2}}-1 \right)}^{\dfrac{1}{2}\times 4}} \\
 & \Rightarrow {{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}=2{{a}^{8}}+12{{a}^{4}}\left( {{a}^{2}}-1 \right)+2{{\left( {{a}^{2}}-1 \right)}^{2}} \\
\end{align}$
On opening the brackets and simplifying, we get
$\begin{align}
  & {{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}=2{{a}^{8}}+12{{a}^{4}}\times {{a}^{2}}-12{{a}^{4}}+2\left( {{\left( {{a}^{2}} \right)}^{2}}-2\times {{a}^{^{2}}}\times 1+{{\left( 1 \right)}^{2}} \right) \\
 & \Rightarrow {{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}=2{{a}^{8}}+12{{a}^{4+2}}-12{{a}^{4}}+2\left( {{a}^{4}}-2{{a}^{2}}+1 \right) \\
 & \Rightarrow {{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}=2{{a}^{8}}+12{{a}^{6}}-12{{a}^{4}}+2{{a}^{4}}-4{{a}^{2}}+2 \\
 & \Rightarrow {{\left( {{a}^{2}}+\sqrt{{{a}^{2}}-1} \right)}^{4}}+{{\left( {{a}^{2}}-\sqrt{{{a}^{2}}-1} \right)}^{4}}=2{{a}^{8}}+12{{a}^{6}}-10{{a}^{4}}-4{{a}^{2}}+2 \\
\end{align}$
This is the required solution.

Note: We can also solve this problem by using the following formulas directly:
$\begin{align}
  & {{\left( x+y \right)}^{4}}={{x}^{4}}+4{{x}^{3}}y+6{{x}^{2}}{{y}^{2}}+4x{{y}^{3}}+{{y}^{4}}......(1) \\
 & {{\left( x-y \right)}^{4}}={{x}^{4}}-4{{x}^{3}}y+6{{x}^{2}}{{y}^{2}}-4x{{y}^{3}}+{{y}^{4}}.......(2) \\
\end{align}$
Now, by adding these two equations we will get:
$\Rightarrow {{\left( x+y \right)}^{4}}+{{\left( x-y \right)}^{4}}=2{{x}^{4}}+16{{x}^{2}}{{y}^{2}}+2{{y}^{4}}$
Now, just substitute $x={{a}^{2}}$ and $y=\sqrt{{{a}^{2}}-1}$
$=2{{\left( {{a}^{2}} \right)}^{4}}+12{{\left( {{a}^{2}} \right)}^{2}}{{\left( \sqrt{{{a}^{2}}-1} \right)}^{2}}+2{{\left( \sqrt{{{a}^{2}}-1} \right)}^{4}}$
Now, apply the Laws of Exponents to simplify the expression. Mainly the Power of a Power Rule, which is ${{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}}$ and Multiplication Rule, which is ${{a}^{x}}\times {{a}^{y}}={{a}^{x+y}}$
$=2{{a}^{2\times 4}}+12{{a}^{2\times 2}}{{\left( {{a}^{2}}-1 \right)}^{\dfrac{1}{2}\times 2}}+2{{\left( {{a}^{2}}-1 \right)}^{\dfrac{1}{2}\times 4}}$
$=2{{a}^{8}}+12{{a}^{4}}\left( {{a}^{2}}-1 \right)+2{{\left( {{a}^{2}}-1 \right)}^{2}}$
$=2{{a}^{8}}+12{{a}^{4}}\times {{a}^{2}}-12{{a}^{4}}+2\left( {{\left( {{a}^{2}} \right)}^{2}}-2\times {{a}^{^{2}}}\times 1+{{\left( 1 \right)}^{2}} \right)$
$=2{{a}^{8}}+12{{a}^{4+2}}-12{{a}^{4}}+2\left( {{a}^{4}}-2{{a}^{2}}+1 \right)$
$\begin{align}
  & =2{{a}^{8}}+12{{a}^{6}}-12{{a}^{4}}+2{{a}^{4}}-4{{a}^{2}}+2 \\
 & =2{{a}^{8}}+12{{a}^{6}}-10{{a}^{4}}-4{{a}^{2}}+2 \\
\end{align}$