Multiple Choice Question (MCQ)
The greatest integer less than or equal to (1+\sqrt{2} )^6

×
196

✓
197

×
198

×
199
The greatest integer less than or equal to (1+\sqrt{2} )^6
196
197
198
199
(1+\sqrt{2} )^6
=[(1+\sqrt{2} )^2]^3
=(3+2\sqrt{2} )^3
Let
Then
According to Vieta's formulas, a,b are two real roots of the quadratic equation
x^26x+1 = 0
Then
a^2 = 6a1
b^2 = 6b1
Using sum of cubes identity and substituting (3), (4), (5) gives
a^3+b^3 = (a+b)(a^2ab+b^2)
=6\times(341)
=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.