<s>[INST] Find $a^3 + b^3 +c^3, $ given $a+b+c=12$ and $a^3 \cdot b^4 \cdot c^5 = 0.1 \cdot (600)^3$ $a+b+c=12$  and  $a^3 \cdot b^4 \cdot c^5 = 0.1 \cdot (600)^3$.
Find  $a^3 + b^3 +c^3 = ?$ 
My approach is to use AM-GM inequality. Is it correct?
 [/INST] If the numbers have to be positive, then the problem can be solved using AM-GM, indeed.
Namely, consider the following 12 numbers: 
$\frac{a}{3}, \frac{a}{3}, \frac{a}{3}, \frac{b}{4}, \frac{b}{4}, \frac{b}{4}, \frac{b}{4}, \frac{c}{5}, \frac{c}{5}, \frac{c}{5}, \frac{c}{5}, \frac{c}{5}$. 
The AM is $1$, as $a+b+c=12$. 
Thus, the GM is at most $1$, so the product of these $12$ numbers is at most $1$, which yields 
$\frac{a^3b^4c^5}{3^3\cdot 4^4\cdot 5^5}\leq 1$. 
As $a^3b^4c^5 = 0.1\cdot 600^3 = 3^3\cdot 4^4\cdot 5^5$, the above inequality is sharp: it holds with equality. 
Thus the AM and the GM of the given $12$ numbers coincide, which means that all $12$ numbers must be equal. 
Hence, $\frac{a}{3}= \frac{b}{4}= \frac{c}{5}$. 
Using $a+b+c=12$ we obtain $a=3, b=4, c=5$. 
Now you just have to sustitute this into $a^3+b^3+c^3$, and you are done. 
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