AIME 2

$\begin{aligned}x^3-xyz&=2\\ y^3-xyz&=6\\ z^3-xyz&=20\\\\ \text{Let, }xyz&=t\\ x^3&=(t+2)\hspace{5mm}\color{#3D99F6}\small(1)\\ y^3&=(t+6)\hspace{5mm}\color{#3D99F6}\small(2)\\ z^3&=(t+20)\hspace{3mm}\color{#3D99F6}\small(3)\\\\ \text{Multiplying,}&\color{#3D99F6}(1),(2)\color{#333333}\text{ and }\color{#3D99F6}(3)\\ (xyz)^3&=(t+2)(t+6)(t+20)\\ t^3&=(t+2)(t+6)(t+20)\hspace{5mm}\color{#3D99F6}( t=xyz)\\ t^3&=t^3+28t^2+172t+240\\ 0&=28t^2+172t+240 \\ 0&=7t^2+43t+60\\ 0&=(7t^2+28t)+(15t+60)\\ 0&=(7t+15)(t+4)\\ \implies t&=-4,\dfrac{-15}{7}\\\\ \text{Adding,}&\color{#3D99F6}(1),(2)\color{#333333}\text{ and }\color{#3D99F6}(3)\\ x^3+y^3+z^3&=28+3t\\ \text{It is clear that}&\text{ in order to maximize the value of } a^3+b^3+c^3 \text{ we have to maximize } t\\ \implies t&=\dfrac{-15}{7}\\ a^3+b^3+c^3&=28+3\cdot\left(\dfrac{-15}{7}\right)\\ &=\dfrac{196-45}{7}\\ &=\dfrac{151}{7}\\ \implies m+n&=151+7=\color{#EC7300}\boxed{\color{#333333}158}\end{aligned}$

Subtracting the first equation from the second:

$\large \displaystyle y^3 - x^3 = 4$

$\color{#20A900} \boxed{ \large \displaystyle x = \sqrt[3]{y^3 - 4} }$

Subtracting the second equation from the third:

$\large \displaystyle z^3 - y^3 = 14$

$\color{#20A900} \boxed{ \large \displaystyle z = \sqrt[3]{ y^3 + 14} }$

Plugging this two equations into the second:

$\large \displaystyle y^3 - \left [ \sqrt[3]{(y^3 - 4)(y^3 + 14)} \right ] y = 6$

$\large \displaystyle y^3 - 6 = \left [ \sqrt[3]{y^6 + 10y^3 -56} \right ] y$

Cubing both sides:

$\large \displaystyle y^9 - 18y^6 + 108y^3 - 216 = y^3 \left ( y^6 + 10y^3 -56 \right )$

$\large \displaystyle y^9 - 18y^6 + 108y^3 - 216 = y^9 + 10y^6 -56y^3$

$\large \displaystyle 28y^6 - 164y^3 + 216 = 0$

$\color{#20A900} \boxed{ \large \displaystyle 7y^6 - 41y^3 + 54 = 0 }$

$\large \displaystyle y^3 = \frac{41 \pm \sqrt{41^2 - 4\cdot 7 \ 54}}{14}$

$\large \displaystyle (i) \ y^3 = \frac{27}{7} \rightarrow x^3 = - \frac{1}{7} \rightarrow \large \displaystyle z^3 = - \frac{125}{7}$

Or:

$\large \displaystyle (ii) \ y^3 = 2 \rightarrow x^3 = -2 \rightarrow z^3 = 16$

$(i)$ leads to:

$\boxed{\large \displaystyle a^3+b^3 + c^3 = \frac{151}{7} }$

While $(ii)$ leads to:

$\boxed{\large \displaystyle a^3+b^3 + c^3 = 16 }$

So, the maximum value is:

$\color{#20A900} \boxed{\large \displaystyle \frac{151}{7}}$

Then:

$\color{#3D99F6} \large \displaystyle m = 151, n = 7, \boxed{\large \displaystyle m + n = 158}$