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# Suppose that $a$, $b$, and $c$ are real numbers for which ​ and a 0 353 3 +598 Suppose that a, b, and c are real numbers for which \begin{align*}a^2b^2 &= 28, \\b^2c^2 &= 21, \text{ and}\\a^2c^2 &= 27,\end{align*} and a michaelcai Oct 12, 2017 edited by michaelcai Oct 12, 2017 edited by michaelcai Oct 12, 2017 edited by michaelcai Oct 12, 2017 #1 +90023 0 What's the question ? CPhill Oct 12, 2017 #2 +598 0 It doesnt show my edits for some reason. Here is the end of the question: and EDIT: UHOQIJKPWHOJKMI IT WONT LET ME USE THE LESS THAN BUTTON here is the end: and \(a Find 3b. EDIT: I HAVE TO WORD IT NOW AHHHGHHHHH ok, so a less than b is less than c. Find 3b. michaelcai Oct 12, 2017 edited by michaelcai Oct 12, 2017 edited by michaelcai Oct 12, 2017 #3 +598 0 Solved it: : Noticing that left-hand sides are symmetric ina$,$b$, and$c$(in that any relabelling of the variables in one of the left-hand sides gives one of the others), we multiply all three equations to get$(a^2b^2c^2)^2=(28)(27)(21)$, which implies$a^2b^2c^2=\sqrt{(28)(27)(21)}=\sqrt{(4\cdot 7)(9\cdot 3)(3\cdot 7.2\cdot 3\dot3\cdot 7=126.$Dividing this equation by each of the three original equations gives$c^2 = 9/2$,$a^2 = 6$, and$b^2=14/3$, respectively. Since$b

michaelcai  Oct 12, 2017