Question

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

Collected in the board: The сube Problems

Steven Zheng posted 1 day ago

Answer

(\sqrt{3} +\sqrt{5} )^6

=[(\sqrt{3} +\sqrt{5})^2]^3

=(8+2\sqrt{15} )^3

Let

a = 8+2\sqrt{15}
(1)
b = 8-2\sqrt{15}
(2)

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
(3)

Using sum of cubes identity and substituting (3), (4), (5) gives

a^3+b^3 = (a+b)(a^2+b^2-ab)

=16\times ( 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


Steven Zheng posted 1 day ago

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