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P=\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}
Rationalize the denominator of \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} by multiplying numerator and denominator by \sqrt{3}-\sqrt{2}.
P=\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}{\left(\sqrt{3}\right)^{2}-\left(\sqrt{2}\right)^{2}}
Consider \left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.
P=\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}{3-2}
Square \sqrt{3}. Square \sqrt{2}.
P=\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}{1}
Subtract 2 from 3 to get 1.
P=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)
Anything divided by one gives itself.
P=\left(\sqrt{3}-\sqrt{2}\right)^{2}
Multiply \sqrt{3}-\sqrt{2} and \sqrt{3}-\sqrt{2} to get \left(\sqrt{3}-\sqrt{2}\right)^{2}.
P=\left(\sqrt{3}\right)^{2}-2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3}-\sqrt{2}\right)^{2}.
P=3-2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^{2}
The square of \sqrt{3} is 3.
P=3-2\sqrt{6}+\left(\sqrt{2}\right)^{2}
To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.
P=3-2\sqrt{6}+2
The square of \sqrt{2} is 2.
P=5-2\sqrt{6}
Add 3 and 2 to get 5.