Question

If $n$ is the degree of the polynomial, ${{\left[ \dfrac{2}{\sqrt{5{{x}^{3}}+1}-\sqrt{5{{x}^{3}}-1}} \right]}^{8}}+{{\left[ \dfrac{2}{\sqrt{5{{x}^{3}}+1}+\sqrt{5{{x}^{3}}-1}} \right]}^{8}}$ and $m$ is the coefficient of ${{x}^{n}}$ in it, then the ordered pair $\left( n,m \right)$ is equal to:
(A). $\left( 12,{{\left( 20 \right)}^{4}} \right)$
(B). $\left( 8,5{{\left( 10 \right)}^{4}} \right)$
(C). $\left( 24,{{\left( 10 \right)}^{8}} \right)$
(D). $\left( 12,8{{\left( 10 \right)}^{4}} \right)$

Answer 1

Hint: We will start by rationalising the terms in the both brackets so that the expression is simplified. Now after we simplify the given expression, we will use the simple binomial theorem to find the coefficient of the asked power in the given expression.

Complete step by step solution:
The given expression is ${{\left[ \dfrac{2}{\sqrt{5{{x}^{3}}+1}-\sqrt{5{{x}^{3}}-1}} \right]}^{8}}+{{\left[ \dfrac{2}{\sqrt{5{{x}^{3}}+1}+\sqrt{5{{x}^{3}}-1}} \right]}^{8}}.$

Rationalise the given polynomial as follows:
\[\Rightarrow {{\left[ \dfrac{2}{\sqrt{5{{x}^{3}}+1}-\sqrt{5{{x}^{3}}-1}} \right]}^{8}}+{{\left[ \dfrac{2}{\sqrt{5{{x}^{3}}+1}+\sqrt{5{{x}^{3}}-1}} \right]}^{8}}\]
\[\Rightarrow {{\left[ \dfrac{2}{\sqrt{5{{x}^{3}}+1}-\sqrt{5{{x}^{3}}-1}}\times \dfrac{\sqrt{5{{x}^{3}}+1}+\sqrt{5{{x}^{3}}-1}}{\sqrt{5{{x}^{3}}+1}+\sqrt{5{{x}^{3}}-1}} \right]}^{8}}+{{\left[ \dfrac{2}{\sqrt{5{{x}^{3}}+1}+\sqrt{5{{x}^{3}}-1}}\times \dfrac{\sqrt{5{{x}^{3}}+1}-\sqrt{5{{x}^{3}}-1}}{\sqrt{5{{x}^{3}}+1}-\sqrt{5{{x}^{3}}-1}} \right]}^{8}}\]
Simplify, it further as follows:
${{\left[ \dfrac{2}{\sqrt{5{{x}^{3}}+1}-\sqrt{5{{x}^{3}}-1}} \right]}^{8}}+{{\left[ \dfrac{2}{\sqrt{5{{x}^{3}}+1}+\sqrt{5{{x}^{3}}-1}} \right]}^{8}}={{\left[ 2\left( \sqrt{5{{x}^{3}}+1}+\sqrt{5{{x}^{3}}-1} \right) \right]}^{8}}+{{\left[ 2\left( \sqrt{5{{x}^{3}}+1}-\sqrt{5{{x}^{3}}-1} \right) \right]}^{8}}$

We can take the term ${{2}^{8}}$ common from both the brackets and write the given expression as follows:
${{\left[ \dfrac{2}{\sqrt{5{{x}^{3}}+1}-\sqrt{5{{x}^{3}}-1}} \right]}^{8}}+{{\left[ \dfrac{2}{\sqrt{5{{x}^{3}}+1}+\sqrt{5{{x}^{3}}-1}} \right]}^{8}}={{2}^{8}}\left[ {{\left( \sqrt{5{{x}^{3}}+1}+\sqrt{5{{x}^{3}}-1} \right)}^{8}}+{{\left( \sqrt{5{{x}^{3}}+1}-\sqrt{5{{x}^{3}}-1} \right)}^{8}} \right]\ldots \text{ }\left( 1 \right)$

The expression in the bracket looks like ${{\left( a+b \right)}^{8}}+{{\left( a-b \right)}^{8}}$ where the term $a=\sqrt{5{{x}^{3}}+1}$ and $b=\sqrt{5{{x}^{3}}-1}.$
Let us consider the binomial expansion for each of the terms separately.
We will start with the first term that is ${{\left( a+b \right)}^{8}}.$
We know that the binomial expression for the above term is given as follows:
${{\left( a+b \right)}^{8}}={}^{8}{{C}_{0}}{{a}^{8}}{{b}^{0}}+{}^{8}{{C}_{1}}{{a}^{7}}{{b}^{1}}+\cdots +{}^{8}{{C}_{8}}{{a}^{0}}{{b}^{8}}\ldots \text{ }\left( 2 \right)$
This is the final expansion we obtained for the first erm.
Similarly, now we will consider the second term that is ${{\left( a-b \right)}^{8}}.$
We know that the binomial expression for it is given as follows:
${{\left( a-b \right)}^{8}}={}^{8}{{C}_{0}}{{a}^{8}}{{b}^{0}}-{}^{8}{{C}_{1}}{{a}^{7}}{{b}^{1}}+\cdots +{}^{8}{{C}_{8}}{{a}^{0}}{{b}^{8}}\ldots \text{ }\left( 3 \right)$
This is the final expression for the second term.
If we add equations (2) and (3) we observe that the even placed terms will get cancelled and we will be left with odd placed terms only.
Therefore, we write the following:
${{\left( a+b \right)}^{8}}+{{\left( a-b \right)}^{8}}=2\left[ {}^{8}{{C}_{0}}{{a}^{8}}{{b}^{0}}+{}^{8}{{C}_{1}}{{a}^{7}}{{b}^{1}}+\cdots +{}^{8}{{C}_{8}}{{a}^{0}}{{b}^{8}} \right]$
Now substituting $a=\sqrt{5{{x}^{3}}+1}$ and $b=\sqrt{5{{x}^{3}}-1}$ in the above equation we write:
\[\Rightarrow {{\left[ \dfrac{2}{\sqrt{5{{x}^{3}}+1}-\sqrt{5{{x}^{3}}-1}} \right]}^{8}}+{{\left[ \dfrac{2}{\sqrt{5{{x}^{3}}+1}+\sqrt{5{{x}^{3}}-1}} \right]}^{8}}\]
\[\Rightarrow {{2}^{8}}\left[ 2\left( \begin{align}
  & {}^{8}{{C}_{0}}{{\left( 5{{x}^{3}}+1 \right)}^{4}}+{}^{8}{{C}_{2}}{{\left( 5{{x}^{3}}+1 \right)}^{3}}{{\left( 5{{x}^{3}}-1 \right)}^{1}}+{}^{8}{{C}_{4}}{{\left( 5{{x}^{3}}+1 \right)}^{2}}{{\left( 5{{x}^{3}}-1 \right)}^{2}}+ \\
 & {}^{8}{{C}_{6}}{{\left( 5{{x}^{3}}+1 \right)}^{1}}{{\left( 5{{x}^{3}}-1 \right)}^{3}}+{}^{8}{{C}_{8}}{{\left( 5{{x}^{3}}-1 \right)}^{4}} \\
\end{align} \right) \right]\]
Now from the above expression it can be clearly seen that the degree of the polynomial is going to be $12$.
Therefore, $n=12$.
Thus, only options A or D can be right.
Now let us find the coefficient of the term ${{x}^{12}}$ using the binomial expansion.
The coefficient is given by:
Coefficient of ${{x}^{12}}=2\left[ {}^{8}{{C}_{0}}{{5}^{4}}+{}^{8}{{C}_{2}}{{5}^{4}}+{}^{8}{{C}_{4}}{{5}^{4}}+{}^{8}{{C}_{6}}{{5}^{4}}+{}^{8}{{C}_{8}}{{5}^{4}} \right]$
Simplify the expression further and obtain the following value for $m$.
$m={{\left( 20 \right)}^{4}}$
Therefore, the ordered pair is given by $\left( n,m \right)=\left( 12,{{\left( 20 \right)}^{4}} \right)$.

So, the correct answer is “Option A”.

Note: Use the method called ‘Root Rationalization of denominator’ which is a process by which radicals in the denominator of an algebraic fraction are eliminated.
For Example: \[\dfrac{1}{\sqrt{2}-3}\] is an irrational denominator so, here by using Rationalization method
So its rationalize form will be \[=\left[ \dfrac{1}{\sqrt{2}-3}\times \dfrac{\sqrt{2}+3}{\sqrt{2}+3} \right]+\left[ \dfrac{1}{\sqrt{2}-3}\times \dfrac{\sqrt{2}-3}{\sqrt{2}-3} \right]\]
Here we have multiplied the numerator and denominator firstly with its conjugate and another the numerator and denominator with real denominator and did the addition of both and simplify it. Rationalization method will help in saving the time of an individual.