Flooring Cubes

The problem is asking the greatest integer that is less than or equal to $(3 + \sqrt{5})^3$ Notice that $(3 - \sqrt{5})$ is between 0 and 1, thus cubing it will lead into a number between 0 and 1, and adding $(3 - \sqrt{5})^3$ which is a small number to $(3 + \sqrt{5})^3$ would not greatly affect the value of $(3 + \sqrt{5})^3$ .

Let $(3 + \sqrt{5})^3 + (3 - \sqrt{5})^3 = S$ . using special factoring, specifically sum of cubes, we can show that $S = 144$ . Since we added (3 - \sqrt{5})^3 which is a number less than 1, then $143 < (3 + \sqrt{5})^3 < 144$ , thus the floor function of $(3 + \sqrt{5})^3 is 143$ . Answer: 143

To get rid of the irrational $\sqrt5$ , I began by adding $(3-\sqrt5)^3$ to $(3+\sqrt5)^3$ . Expanding both of the parts yields $(27+27\sqrt5+45+5\sqrt5)$ for $(3+\sqrt5)$ and $(27-27\sqrt5+45-5\sqrt5)$ for $(3-\sqrt5)$ . Adding these together cancels the terms containing multiples of $\sqrt5$ . This results in 144. We must now subtract the approximate quantity of $(3-\sqrt5)^3$ to find the value that is asked for. Noting that $0<(3-\sqrt5)<1$ allows us to find that cubing this will result in a number between 0 and 1. Subtracting any value less than 1, but greater than 0 will yield the result 143 with some sort of decimal attached to it. Evaluating the floor function of this gives the final result of 143.

$(x+y)^3 = x^3+x^2y+xy^2+y^3$

$(3+\sqrt{5})^3 = 3^3+3^2\cdot\sqrt{5}+3\cdot(\sqrt{5})^2+(\sqrt{5})^3$

$= 27+27\sqrt{5}+45+5\sqrt{5}$

$= 72+32\sqrt{5}$

[ $\sqrt{5} = 2.236$ ]

$(3+\sqrt{5})^3 = 72+71.55 = 143.55$

$|(3+\sqrt{5})^3| = 143$

By (a+b)^3= a^3 + b^3 + 3.a^2.b + 3.a.b^2.

We get, [(3+ \sqrt{5})^3]= [3^3 + sqrt{5}^3 + 3.3^2.sqrt{5} + 3.3.sqrt{5}^2]

=> [27+ 5.sqrt{5] + 27.sqrt{5} + 9.5]

=> [27+45+32.sqrt{5}]

=> [72+ 32.(2.236) ] [ Since, sqrt{5} = 2.236 (approx.) ]

=> [72 + 71.55]

=> [143.55]

By the greatest integer function, the answer is 143.

Let R = $(3+\sqrt{5})^3 = I + f$ , where $I = \lfloor R \rfloor$ , and $0\le f<1$ .

Let $(3-\sqrt{5})^3$ = f', where $0< f'<1$ (as $0<(3-\sqrt{5}) < 1$ )

Now, $I+f+f' = (3+\sqrt{5})^3 + (3-\sqrt{5})^3$

On the RHS, the irrational terms will cancel out to give an integer, making I + f + f' an integer. Since I is an integer, therefore f + f' must also be an integer

As 0< f'<1 and $0\le f<1$ , therefore 0<f+f'<2. The only possible integral value of f+f' is 1.

Therefore we have I + 1 = $(3+\sqrt{5})^3 + (3-\sqrt{5})^3$ = 144, making I = 143

\lfloor (3 + \sqrt{5})^3 \rfloor Expand the brackets. Therefore, \lfloor (72+32\sqrt{5}) \rfloor Finding the square root of 5 without calculator. A video illustrating the method can be found here. http://www.youtube.com/watch?v=3i94NWF39nU The square root of 5 has to only be to 3 decimal places. After it has been found simpely carry out the operation and you should get an answer similar to 143.552. Forget about the decimals and you should get an answer of 143.

Value of √5 = 2.236. 3 + √5 = 5.236 (5.236)^{3} = 143.548......... Since we have to apply the greatest integer function and the number is positive, it is not needed to consider the decimal part. [143.548.........] = 143.

Consider the Conjugate of $3 + \sqrt{5}$ , which is $3 - \sqrt{5}$ .

By expanding, we can show that $(3 + \sqrt{5})^3 + (3 - \sqrt{5})^3 = 2( 3^3 + 3 \times 3 \times 5) = 144$ .

We will use the fact that $0 < 3 - \sqrt{5} < 1$ , so $0 < (3- \sqrt{5})^3 < 1$ . Hence, $\left \lfloor ( 3 + \sqrt{5} ) ^3 \right \rfloor = 143$ .

Expanded value of (3 + sqrt 5)^3 is 72 + 32 sqrt 5 which is approximately143.55. So required value is 143.

Greatest Integer Function: ⌊x⌋:R→Z refers to the greatest integer less than or equal to x. For example ⌊2.3⌋=2 and ⌊−3.4⌋=−4.

expanding the bracket by $(a+b)^3=a^3+b^3+3a*b(a+b)$

(3+√5)^3 =⌊72+32* $\sqrt{5}$ ⌋

hence the given equals ⌊72+32* $\sqrt{5}$ ⌋=⌊143.551⌋=143

Using the trinomial expansion of the cubic expression, I got 27 + 27 $\sqrt{5}$ +45 + 5 $\sqrt{5}$ = 72 + 32 $\sqrt{5}$ . Using $\sqrt{5}$ ~ 2.23, I had a sum of a little over 143. So the answer would be 143

On expanding we get

=27 + 27(5)^1/2 + 45 + 5(5)^1/2

=72 + 32(5)^1/2

=143.552

So the solution is 143

First, we expand $(3 + \sqrt{5} )^3$ . We can do this by using binomial expansion.
${3 \choose 3} 3^3\sqrt{5}^0 + {3 \choose 2} 3^2\sqrt{5}^1 + {3 \choose 1} 3^1\sqrt5^2 + {3 \choose 0} 3^0\sqrt{5}^3\\ 1(27)(1) + 3(9)(\sqrt(5)) + 3(3)(5) + 1 (1)(5\sqrt{5})\\ 72 + 32\sqrt{5}$

We can say that $\sqrt{5} \approx 2.24$ because everything beyond the hundedths place is negligible in dealing with floors. Therefore we can use $72 + 32(2.23) = 72 + 71.36 = 143.36$
And since we are dealing with floor values, we say $\lfloor143.36\rfloor = 143$ .

I saw you require the fact of $0 < 3 - \sqrt{5} < 1$ which cannot be known without calculator (at least to me) , so I have another approximation to the answer :

Expanding $( 3+\sqrt{5})^3$ ,

we get $72 + 32\sqrt{5}$ ,by putting $4$ from $32$ into the square root ,

we get $72+8\sqrt{80}$ , obviously $\sqrt{81} > \sqrt{80}$ ,

and the square root approximation( $\sqrt{n+\delta x}$ ) has less error with larger $n$ and smaller $\delta x$ so i chose $80 - 81$ combination. (for reference , error = $0.0557$ )

(other combination available will be $320 - 324$ (error = $0.1114$ ,etc but for simplicity I used $80 - 81$ .

So $72 + 8\sqrt{81} = 144 > ( 3+\sqrt{5})^3$ and $\lfloor (3+\sqrt{5})^3 \rfloor = \boxed{143}$