3, 2, 1 blast off

$\begin{aligned} \displaystyle {3 \sqrt[3]{\sqrt[3]{2} - 1}} &= 3 \sqrt[3]{\frac{(\sqrt[3]{2} - 1)(\sqrt[3]{2^2} + \sqrt[3]{2 \times 1} + \sqrt[3]{1})}{\sqrt[3]{2^2} + \sqrt[3]{2 \times 1} + \sqrt[3]{1}}}\\ &= \frac{3}{\sqrt[3]{\sqrt[3]{4} + \sqrt[3]{2} + 1}}\\ &= \frac{3 \sqrt[3]{3}}{\sqrt[3]{2 + 3\sqrt[3]{4} + 3\sqrt[3]{2} + 1}}\\ &= \frac{3 \sqrt[3]{3}}{\sqrt[3]{2} + 1}\\ &= \frac{3 \sqrt[3]{3}(\sqrt[3]{2^2} - \sqrt[3]{2 \times 1} + \sqrt[3]{1})}{(\sqrt[3]{2} + 1)(\sqrt[3]{2^2} - \sqrt[3]{2 \times 1} + \sqrt[3]{1})}\\ &= \frac{3 \sqrt[3]{3}(\sqrt[3]{4} - \sqrt[3]{2} + \sqrt[3]{1})}{3}\\ &= \sqrt[3]{12} + \sqrt[3]{3} - \sqrt[3]{6} \end{aligned}$ Hence $a + b + c = 12 + 3 + 6 = 21$

Ok I will try to write my solution as neat as possible.

Aim:

In this problem, we should try to do something on the $\text{LHS}$ and not the $\text{RHS}$ or both. The reason is because we are not solving an equation, how can we possible solve for three variables with one equation? Hence, we should try to arrange $\text{LHS}$ into the form of $\text{RHS}$ . For example

$3+4\sqrt7=a+b\sqrt7$ , solve for $a$ and $b$

Immediately we can see $a=3$ and $b=4$ . This is because we analyse the form or pattern of both expression. So, solving this kind of problems need lots of creativity!

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Firstly, we set some variables to see the connection between $3,2,1$ . Let

$x^3=2$ , $x=\sqrt[3]{2}$

$y^3=1$ , $y=1$

Now, we substitute it to the expression

$3\sqrt[3]{\strut\sqrt[3]{2}-1}$

$=(x^3+y^3)\sqrt[3]{\strut x-y}$

$=(x+y)(x^2-xy+y^2)\sqrt[3]{\strut x-y}$

$=(x^2-xy+y^2)\sqrt[3]{\strut(x-y)(x+y)^3}$

$=(x^2-xy+y^2)\sqrt[3]{\strut x^4+2x^3y-2xy^3-y^4}$

Now, we substitute all the numbers back into the expression we have

$(\sqrt[3]{4}-\sqrt[3]{2}+1)\sqrt[3]{\strut 2\sqrt[3]{2}+4-2\sqrt[3]{2}-1}$

$=(\sqrt[3]{4}-\sqrt[3]{2}+1)\sqrt[3]{3}$

$=\sqrt[3]{12}-\sqrt[3]{6}+\sqrt[3]{3}$

Let's arrange it to the form same as the problem

$\sqrt[3]{12}+\sqrt[3]{3}-\sqrt[3]{6}$

Hence, we can see $a=12$ , $b=3$ , $c=6$ and $a+b+c=21$

If $\alpha = 2^{\frac13}$ , then $1 = \alpha^3-1 = (\alpha-1)(\alpha^2+\alpha+1)$ and $3 = \alpha^3 + 1 = (\alpha+1)(\alpha^2-\alpha+1)$ . Thus $\frac{3}{\alpha-1} \; = \; 3(\alpha^2 + \alpha + 1) \; = \; \alpha^3 + 3\alpha^2 + 3\alpha +1 \; = \; (\alpha + 1)^3$ and so $3(\alpha-1)^{\frac13} \; = \; \frac{3^{\frac43}}{\alpha+1} \; = \; 3^{\frac13}(\alpha^2 - \alpha + 1)$ From this we can readily spot the solution $a = 12$ , $b = 3$ , $c = 6$ , and hence $a+b+c=21$ .

How to show that this is the only solution is more of a challenge. It would seem likely that we need to show that $a,b,c$ are all multiples of $3$ and that $(a/3)^{\frac13},(b/3)^{\frac13},(c/3)^{\frac13}$ all belong to $\mathbb{Z}(\alpha)$ , but I have not managed that yet. A computer search determines that there are no other solutions within the range of Brilliant solutions!

use your calculator and consider 1.9^3 as 7 therefore 7+7-7=21 and therefore a,b,c are 7,7,7 hence A+B+C=21