Guilherme's algebraic expression

We see that

$45+29\sqrt{2}=27+18+27\sqrt{2}+2\sqrt{2}$ $=3^3+(\sqrt{2})^3+3\cdot(3)\cdot(\sqrt{2})^2+3\cdot(3)^2\cdot(\sqrt{2})$

$\Rightarrow 45+29\sqrt{2}=(3+\sqrt{2})^3$

Similarly, $45-29\sqrt{2}=(3-\sqrt{2})^3$

Therefore, $A=\sqrt{24(3-\sqrt{2}+3+\sqrt{2})}=\sqrt{24 \times 6}=\sqrt{144}=12$

Since the most complicated part is $\sqrt[3]{45 - 29\sqrt{2}} + \sqrt[3]{45 + 29\sqrt{2}}$ , let's simplify that part first.

Let $X = \sqrt[3]{45 - 29\sqrt{2}}$ and $Y = \sqrt[3]{45 + 29\sqrt{2}}$ .

There is a well-known identity, which is:

$X^3+ Y^3 = (X + Y) (X^2 - XY + Y^2)$

If we rearrange the identity, it becomes:

$X + Y = \frac{X^3+ Y^3} {X^2 - XY + Y^2}$

When we substitute $X = \sqrt[3]{45 - 29\sqrt{2}}$ and $Y = \sqrt[3]{45 + 29\sqrt{2}}$ , it becomes:

$X + Y = \frac{(\sqrt[3]{45 - 29\sqrt{2}})^3+ (\sqrt[3]{45 + 29\sqrt{2}})^3} {X^2 - XY + Y^2}$

$X + Y = \frac{45 - 29\sqrt{2}+ 45 + 29\sqrt{2}} {X^2 - XY + Y^2}$

$X + Y = \frac{90} {X^2 - XY + Y^2}$ ----------(1)

Take note that:

$XY = (\sqrt[3]{45 - 29\sqrt{2}})(\sqrt[3]{45 + 29\sqrt{2}})$

$XY = \sqrt[3]{(45 - 29\sqrt{2})(45 + 29\sqrt{2})}$

According to the identity $(a - b)(a + b) = a^2 - b^2$ ,

$XY = \sqrt[3]{45^2 - (29\sqrt{2})^2}$

$XY = \sqrt[3]{2025 - 1682}$

$XY = \sqrt[3]{343}$

$XY = 7$ ----------(2)

In addition, since $X^2 + Y^2 = (X + Y)^2 - 2XY$ ,

$X^2 + Y^2 = (X + Y)^2 - (2)(7)$

$X^2 + Y^2 = (X + Y)^2 - 14$ ----------(3)

Substitute equation (2) and (3) into (1):

$X + Y = \frac{90} {(X + Y)^2 - 14 - 7}$

$X + Y = \frac{90} {(X + Y)^2 - 21}$

Manipulating the equation yields:

$(X + Y)^3 -21(X+Y) -90 = 0$

After factorising the solution, we have:

$((X + Y)-6) ((X + Y)^2 + 6(X+Y) + 15) = 0$

and the only real root is $X + Y = 6$ .

Substitute $X + Y = 6$ into the main equation:

$A = \sqrt{24 (X+Y)}$

$A = \sqrt{24 (6)}$

$A = \sqrt{144}$

Hence, $A = \boxed {12}$ .

This is a beautiful expression! However, it is too complicated.

So, let us set some variables and see what would it become

$45-29\sqrt{2}\equiv(a-b)^3----------(1)$

$45+29\sqrt{2}\equiv(a+b)^3----------(2)$

Then we will have:

$A=\sqrt{24(\sqrt[3]{(a-b)^3}+\sqrt[3]{(a+b)^3})}$

$A=\sqrt{24((a-b)+(a+b))}$

$A=\sqrt{24(2a)}$

$A=\sqrt{48a}$

Interestingly, the expression is now more "friendly", isn't it? So all we need to do is solve for $a$ !

$(1)+(2)$ , we will get:

$a^3+3ab^2=45----------(4)$

$(1)-(2)$ , we will get:

$b^3+3a^2b=29\sqrt{2}----------(5)$

There is no $\sqrt{2}$ in $(4)$ but it is present in $(5)$ , so we can conclude that $\sqrt{2}$ is squared in $(4)$ , which is $b$ and we can get $b=\sqrt{2}$

Substitute $b=\sqrt{2}$ into $(5)$ and we will get:

$2\sqrt{2}+3\sqrt{2}a^2=29\sqrt{2}$

$a=\pm3$

However in this case, $a$ is a positive number for $A$ to be real. So we conclude

$a=3$

Finally, substitute $a=3$ into the previous "friendly" expression and we will get

$A=12$

The problem is fairly simple to a mathematical intuitionist. Since cube roots are appearing, one can easily suspect that the quantity inside the cube root must be a perfect cube. By a little bit of analysis, you find out that the quantity in the first cube root is a perfect cube of (3-√2) and the quantity inside the second cube root is a perfect cube of (3+ √2). Now the problem becomes quite simple. On adding both the quantities, we get square root of (24*6), which is square root of 144, i.e. 12.

take the cube root of 45-29root 2 as (x-y) and the other as x+y multiply both with each other and then cube the first equation by subtituition u will get x and then rest u can understand

We realize that the first cube root expression is of the form $a+b\sqrt{2}$ , and the second expression of the form $a-b\sqrt{2}$ . So the entire expression becomes $\sqrt{24(2a)}$ . Evaluating at $a=1,2$ doesn't result in an integer, but $a=3$ does, which makes the answer $\boxed{12}$ .

Note: the next highest a that works is a=12, which is unrealistic.

If you're smart enough, use calculator. XD Sorry. That's what I just did. :3