The integral part of \[{\left( {\sqrt 2 + 1} \right)^6}\] is
A. 197
B. 196
C. 175
D. 176

Hint: First of all, consider the given binomial as the sum of integral part and fractional part. Then add up this binomial with its contemporary binomial to find the integral part of the given binomial, use the binomial theorem for the expansion of the binomials. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Let the \[{\left( {\sqrt 2 + 1} \right)^6} = I + F\]
Where, \[I\] is an integer and \[F\] is a fractional part i.e., \[0 < F < 1\]
Let \[{\left( {\sqrt 2 - 1} \right)^6} = f\]
Where, \[f\] is a fractional part i.e., \[0 < f < 1\] and hence, \[0 < {\left( {\sqrt 2 - 1} \right)^6} < 1\].
We know that \[{\left( {x + y} \right)^n} = {}^n{C_0}{x^n} + {}^n{C_1}{x^{n - 1}}y + {}^n{C_2}{x^{n - 2}}{y^2} + ........... + {}^n{C_r}{x^{n - r}}{y^r}................ + {}^n{C_n}{y^n}\] and \[{\left( {x - y} \right)^n} = {}^n{C_0}{x^n} + {\left( { - 1} \right)^1}{}^n{C_1}{x^{n - 1}}y + {\left( { - 1} \right)^2}{}^n{C_2}{x^{n - 2}}{y^2} + ........... + {\left( { - 1} \right)^r}{}^n{C_r}{x^{n - r}}{y^r}................. + {\left( { - 1} \right)^n}{}^n{C_n}{y^n}\]
Consider,
\[
   \Rightarrow {\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6} = \left[ {{}^6{C_0}{{\left( {\sqrt 2 } \right)}^6} + {}^6{C_1}{{\left( {\sqrt 2 } \right)}^5}{{\left( 1 \right)}^1} + {}^6{C_2}{{\left( {\sqrt 2 } \right)}^4}{{\left( 1 \right)}^2} + {}^6{C_3}{{\left( {\sqrt 2 } \right)}^3}{{\left( 1 \right)}^3} + {}^6{C_4}{{\left( {\sqrt 2 } \right)}^2}{{\left( 1 \right)}^4} + {}^6{C_5}{{\left( {\sqrt 2 } \right)}^1}{{\left( 1 \right)}^5} + {}^6{C_6}{{\left( 1 \right)}^6}} \right] - \\
  {\text{ }}\left[ {{}^6{C_0}{{\left( {\sqrt 2 } \right)}^6} - {}^6{C_1}{{\left( {\sqrt 2 } \right)}^5}{{\left( 1 \right)}^1} - {}^6{C_2}{{\left( {\sqrt 2 } \right)}^4}{{\left( 1 \right)}^2} - {}^6{C_3}{{\left( {\sqrt 2 } \right)}^3}{{\left( 1 \right)}^3} - {}^6{C_4}{{\left( {\sqrt 2 } \right)}^2}{{\left( 1 \right)}^4} - {}^6{C_5}{{\left( {\sqrt 2 } \right)}^1}{{\left( 1 \right)}^5} - {}^6{C_6}{{\left( 1 \right)}^6}} \right] \\
\]
Cancelling the common terms, we get
\[
   \Rightarrow {\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6} = 2\left[ {{}^6{C_0}{{\left( {\sqrt 2 } \right)}^6} + {}^6{C_2}{{\left( {\sqrt 2 } \right)}^4}{{\left( 1 \right)}^2} + {}^6{C_4}{{\left( {\sqrt 2 } \right)}^2}{{\left( 1 \right)}^4} + {}^6{C_6}{{\left( 1 \right)}^6}} \right] \\
   \Rightarrow {\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6} = 2\left[ {{}^6{C_0} \times 8 + {}^6{C_2} \times 4 + {}^6{C_4} \times 2 + {}^6{C_6} \times 1} \right] \\
   \Rightarrow {\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6} = 2\left[ {1 \times 8 + 15 \times 4 + {}^6{C_4} \times 2 + 1 \times 1} \right] \\
   \Rightarrow {\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6} = 2\left[ {99} \right] \\
   \Rightarrow {\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6} = 198{\text{ }} \\
  {\text{ }} \\
\]
But we have \[{\left( {\sqrt 2 + 1} \right)^6} + {\left( {\sqrt 2 - 1} \right)^6} = I + F + f\]. So, we get
\[
   \Rightarrow I + F + f = 198 \\
  {\text{ }} \\
\]
Here \[I\]is an integer and 198 is an integer. And \[I + F + f = 198\] is only possible when \[F + f\] is an integer.
But we have \[0 < F < 1\] and \[0 < f < 1\]. Adding up them, we get
\[
   \Rightarrow 0 + 0 < F + f < 1 + 1 \\
   \Rightarrow 0 < F + f < 2 \\
\]
We know that the only 1 is the integer which is greater than 0 and lesser than 2. So, the value of \[F + f = 1\].
Therefore, we get
\[
   \Rightarrow I + 1 = 198 \\
  \therefore I = 198 - 1 = 197 \\
\]
Hence, the integral part of \[{\left( {\sqrt 2 + 1} \right)^6}\] is 197.
Thus, the correct option is A. 197

Note: Here we have added \[{\left( {\sqrt 2 - 1} \right)^6}\]to the given binomial since its value is less than one and greater than zero i.e., \[0 < {\left( {\sqrt 2 - 1} \right)^6} < 1\]. Remember this method for finding the integral parts of the binomials.