Suppose that a, b, and c are real numbers for which \(\begin{align*} a^2b^2 &= 28, \\ b^2c^2 &= 21,\\ a^2c^2 &= 27, \end{align*}\)and a
+600 I'll solve for $a,b,c$?
Multiplying them all together, $a^4b^4c^4=2^2\cdot 3^4\cdot 7^2\implies a^2b^2c^2=2\cdot 9\cdot 7$. So $c^2=\frac{a^2b^2c^2}{a^2b^2}=\frac{9}{2}$, now plug into eqn 3 to find a and then plug into eqn 1 to find b.
