I think min $-4$, Max$2$

in fact, we can find all solutions of the given conditions since it is equivalent to \[(a+b-2)\left( \left( \frac{a+b}{2}+2\right) ^{2}+\frac{3}{4}(a-b)^{2}\right)=0\]

If $a+b$ can take some value $k$ for $a,\ b$ such that $a^3+b^3=8-6ab$, then there exist some real number $(a,\ b)$ satisfying the system of euation : $a+b=k$ and $a^3+b^3=8-6ab$. Elliminating $b$ gives $3(k-2)a^2-3k(k-2)a+k^3-8=0\Longleftrightarrow (k-2)(3a^2-3ka+k^2+2k+4)=0$ When $k\neq 2$, we have $3a^2-3ka+k^2+2k+4=0\ \cdots (*)$, the condition for which there exists some real number $a$, $D=9k^2-12(k^2+2k+4)\geq 0\Longleftrightarrow (k+4)^2\leq 0$, yielding $k=-4$. Therefore the range of $a+b$ is $-4\ or\ 2$. The desired answer is the maximum value $2$ when $(a,\ b)=(t,\ t-2)$ for some real number $t$, the minimum value is $-4$ when $(a,\ b)=(-2,-2)$. Comment: We can factor $a^3+b^3-8+6ab=0$ $\Longleftrightarrow (a+b-2)(a^2+b^2+4-ab+2a+2b)=0$ $\Longleftrightarrow \frac 12(a+b-2)\{(a-b)^2+(a+2)^2+(b+2)^2\}=0$, yielding the line $a+b=2$ or the point $(a,\ b)=(-2,-2)$ which is explained by Kimpul.

\[ a^{3}+b^{3}=8-6ab. \] has the form $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz)$ where $x = a$, $y = b$ and $c = -2$. That's probably how Kimpul came up with the product of those two terms.

So, another of those understating questions; the given condition, equivalent to \[(a+b-2)\left( \left( \frac{a+b}{2}+2\right) ^{2}+\frac{3}{4}(a-b)^{2}\right)=0,\] has as solutions $a+b=2$ or then $\frac{a+b}{2}+2 = 0$ and $(a-b)^{2} = 0$, thus $a=b=-2$, for which $a+b = -4$. Thus the quantity $a+b$ can only take two values, namely $2$ and $-4$. Quite a tall order, to ask for the maximum and minimum value(s) ...

Generalization $a,b,k$ are real numbers such that: $ a^{3}+b^{3}=k^{3}-3kab. $ Find the maximal and minimal value of$a+b$. Ion Bursuc

k＞0, Max. k, min. -2k k＜0, Max. -2k, min. k k=0, a+b=0

Let $a=x, b=y, -2 = z$ Therefore, we get: $x^3+y^3= -z^3+3xyz \implies x^3+y^3+z^3-3xyz= 0$ Taking little cases, we can easily guess that $x+y+z$ is a factor of this. Luckily, the intuition turns out to be true and we get: $x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-xz)$ The result follows.