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Find all solutions to the system of equations:

$$abc^3=24$$ $$ab^3c=54$$ $$a^3bc=6$$

Well by guessing or trial and error, I was able to find

$a=1, b=3, c=2$

but it said find all solutions, so there is a possbility that there could be other solutions.

If $a=1, b=3, c=2$ is the only solutions, how would you solve the question more systematically instead of trial and error.

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There's one common factor in all of them: $abc$. Isolating it, you have $$abc=\frac{24}{c^2}=\frac{54}{b^2}=\frac{6}{a^2}$$ Breaking this up, you get $$\frac{24}{c^2}=\frac{6}{a^2}\to c=2a$$ $$\frac{54}{b^2}=\frac{6}{a^2}\to b=3a$$ You should be able to solve for all three variables now.

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Another approach: multiply all left and right hand sides: $$a^5b^5c^5=7776=6^5.$$ This yields $abc=6$. Plugging this into the equations gives $c^2=4,b^2=9,a^2=1$. Hence the solution triples $(a,b,c)$ are $(1,3,2), (1,-3,-2),(-1,3,-2)$ and $(-1,-3,2)$.

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