Find the greatest integer not greater than $$(\sqrt{3} +\sqrt{5} )^6$$

Question

Find the greatest integer not greater than $$(\sqrt{3} +\sqrt{5}   )^6$$

Uniteasy Admin · Accepted Answer

$$(\sqrt{3} +\sqrt{5}  )^6$$
$$=[(\sqrt{3} +\sqrt{5})^2]^3 $$
$$=(8+2\sqrt{15} )^3$$
Let 
$$a = 8+2\sqrt{15}$$n
$$b = 8-2\sqrt{15}$$n
Then
$$a+b = 16$$
$$ab = 4$$
According to Vieta's formulas, $$a,b $$ are two real roots of the quadratic equation
$$x^2-16x+4=0$$
Then
$$a^2 = 16a-4$$
$$b^2 = 16b-4$$
$$a^2+b^2 = 16(a+b)-8=248$$n
Using sum of cubes identity and substituting (3), (4), (5) gives
$$a^3+b^3 = (a+b)(a^2+b^2-ab)$$
$$=16	imes ( 248-4) =  3904 $$ 
Since 
$$0< b<1$$
then
$$0< b^3<1$$
$$a^3 = 3904-b^3$$
Therefore
The greatest integer that is less than $$a^3$$, that is $$(\sqrt{3} +\sqrt{5}  )^6$$, is 3903

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