The greatest integer less than or equal to $$(1+\sqrt{2} )^6$$

Question

The greatest integer less than or equal to $$(1+\sqrt{2}  )^6$$

Uniteasy Admin · Accepted Answer

$$(1+\sqrt{2}  )^6$$
$$=[(1+\sqrt{2}  )^2]^3$$
$$=(3+2\sqrt{2}  )^3$$
Let
$$a=3+2\sqrt{2}$$n
$$b=3-2\sqrt{2}$$n
Then
$$a+b=6$$n
$$ab = 1$$n
According to Vieta's formulas, $$a,b$$ are two real roots of the quadratic equation
$$x^2-6x+1 = 0$$
Then
$$a^2 = 6a-1$$
$$b^2 = 6b-1$$
$$a^2+b^2 = 6(a+b)-2 = 34$$n
Using sum of cubes identity and substituting (3), (4), (5) gives
$$a^3+b^3 = (a+b)(a^2-ab+b^2)$$
$$=6	imes(34-1)$$
$$=198$$
Therefore,
$$a^3 = 198 - b^3$$
Since $$0< b< 1, 0< b^3< 1$$
The greatest integer less than $$a^3$$ is 197
B is the correct answer.

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