+619 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
+619 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.
+619 Solved it: : Noticing that left-hand sides are symmetric in $a$, $b$, 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
