Find $[(\sqrt 2 +1)^8]$, where $[.]$ is integer part function

Question

$$2^4 (1+\frac{1}{\sqrt 2})^8$$ $$=16 (1+8\frac{1}{\sqrt 2})$$ $$=106.4$$ which is nowhere close to the actual answer. What’s wrong with my approximation?

Answer 1

$(1+x)^n$ can be approximated to $1+nx$ only when $x \to 0$
So your method will not work

Let's add $(1 - \sqrt{2})^8$ on both sides

$(1+\sqrt{2})^8 + (1 - \sqrt{2})^8 = 2(1+ \binom{8}{2}2+\binom{8}{4}4+\binom{8}{6}8+16) = 1154$

$(1+\sqrt{2})^8 + (1 - \sqrt{2})^8 = 1154$

$(1+\sqrt{2})^8 = 1154-(1 - \sqrt{2})^8$

$1154$ is an integer and $(1 - \sqrt{2})^8$ is a very small fraction.

Therefore,

$[(1+\sqrt{2})^8] = 1154-1 = 1153$