Question

Find ${\left( {x + 1} \right)^6} + {\left( {x - 1} \right)^6}$. Hence or otherwise evaluate ${\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6}$.

Answer 1

Hint: In order to solve this question, we have to expand the given expression. Since, we do not have the formula to expand this expression to the power of 6, so we will use the Binomial theorem to solve the expression.


Formula Used:
According to the binomial theorem-
For a polynomial ${\left( {x + 1} \right)^6}$ the Binomial expansion is given by the formula-
${\left( {1 + x} \right)^n} = \sum\limits_{k = 0}^n {{}^n{C_k}{x^k}} $
Where, the value of ${}^n{C_k} = \dfrac{{n!}}{{k!\left( {n - k} \right)!}}$ and $n$ is a positive integer.
We know the number of terms in the expansion would be $n + 1$.


Complete step by step solution
Given:
The polynomial given is ${\left( {x + 1} \right)^6} + {\left( {x - 1} \right)^6}$.
First, we have to expand this then from this expression we have to evaluate,
${\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6}$
For the first part of the expression we will use binomial expansion to solve, we get,\[{\left( {x + 1} \right)^6} = {}^6{C_0}{x^6} + {}^6{C_1}{x^5} + {}^6{C_2}{x^4} + {}^6{C_3}{x^3} + {}^6{C_4}{x^2} + {}^6{C_5}{x^1} + {}^6{C_6}{x^0}\]
Similarly, for the second part we will again use binomial expansion to solve, we get,\[{\left( {x - 1} \right)^6} = {}^6{C_0}{x^6} - {}^6{C_1}{x^5} + {}^6{C_2}{x^4} - {}^6{C_3}{x^3} + {}^6{C_4}{x^2} - {}^6{C_5}{x^1} + {}^6{C_6}{x^0}\]
Now, on combining both parts we get,
\[{\left( {x + 1} \right)^6} + {\left( {x - 1} \right)^6} = 2\left[ {{}^6{C_0}{x^6} + {}^6{C_2}{x^4} + {}^6{C_4}{x^2} + {}^6{C_6}{x^0}} \right]\]
On further solving the combination \[{}^6{C_0} = 1\] and \[{}^6{C_6} = 1\] we get, \[\begin{array}{c}
{\left( {x + 1} \right)^6} + {\left( {x - 1} \right)^6} = 2\left[ {1 \times {x^6} + 15 \times {x^4} + 15 \times {x^2} + 1 \times {x^0}} \right]\\
{\left( {x + 1} \right)^6} + {\left( {x - 1} \right)^6} = 2\left[ {{x^6} + 15{x^4} + 15{x^2} + 1} \right]
\end{array}\]

Now, we know that this is the generalized form of the expression and using this form of expression we can easily evaluate ${\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6}$.
So, if we put $x = \sqrt 2 $ in the expression above, we get,\[\begin{array}{c}
{\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6} = 2\left[ {{{\left( {\sqrt 2 } \right)}^6} + 15{{\left( {\sqrt 2 } \right)}^4} + 15{{\left( {\sqrt 2 } \right)}^2} + 1} \right]\\
 = 2\left[ {\left( 8 \right) + 15\left( 4 \right) + 15\left( 2 \right) + 1} \right]\\
 = 2\left[ {8 + 60 + 30 + 1} \right]\\
 = 198
\end{array}\]

Therefore, the value of the expression ${\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6}$ calculated is 198.


Note: For the expansion of the binomial having larger power we always use the Binomial theorem, but as the power of the binomial increases the harder this method gets. For example, the expansion of the binomial in the given question was easy because the power of the binomial was 6. That is the number of the terms in the expansion would be 7, which is fairly easy to manage.