<s>[INST] If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4,$ then $(a-b)^2=\;$? If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4$, what is the value of $(a-b)^2$?  I think $a^2+b^2=36$, please confirm and is it possible to to figure out one of the variables?
 [/INST] Hints:
$\frac{1}{a^2}+\frac{1}{b^2} = 4 \rightarrow b^2 + a^2 = 4(ab)^2$
$(a-b)^2 = (a^2+b^2) -2(ab)$
edit:
To solve for a particular variable, you can use $ab=3 \rightarrow a = \frac{3}{b}$ to eliminate a variable. For example $a^2+b^2 = \frac{9}{b^2}+b^2$
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