Help with Algebra 2B? Rationalize the denominator and simplify (wait for the equation in comments):

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Help with Algebra 2B? Rationalize the denominator and simplify (wait for the equation in comments):

Mathematics
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\[\frac{ 3\sqrt{2}-2\sqrt{3} }{ 3\sqrt{2}+2\sqrt{3} }\]
Multiply top and bottom by the conjugate.

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So @Hero it would be \[\frac{ (3\sqrt{2}-2\sqrt{3})(3\sqrt{2}-2\sqrt{3}) }{ (3\sqrt{2}+2\sqrt{3})(3\sqrt{2}-2\sqrt{3}) }\] which would simplify to \[\frac{ 3\sqrt{2}^{2}+2\sqrt{3}^{2} }{ 3\sqrt{2}^{2}-2\sqrt{3}^{2} }\] am I right so far?
It's more like: \[\frac{ (3\sqrt{2}-2\sqrt{3})(3\sqrt{2}-2\sqrt{3}) }{ (3\sqrt{2}+2\sqrt{3})(3\sqrt{2}-2\sqrt{3}) } = \frac{(3\sqrt{2}-2\sqrt{3})^2}{(3\sqrt{2})^2 - (2\sqrt{3})^2}\]
Oh ok thanks that helps a lot I think I can do it from here.

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