Answer:
[tex]\dfrac{4\sqrt{3b}+4b\sqrt{b}}{3-b^2}[/tex]
Step-by-step explanation:
The factoring of the difference of squares is used to "rationalize the denominator" of rational functions having square roots in the denominator.
The factoring of the difference of squares is ...
a² -b² = (a -b)(a +b)
This means that an expression in which 'a' or 'b' or both are square roots can be made "rational" by making use of this relation.
To eliminate the radical from the denominator, we multiply numerator and denominator by the "conjugate" of the denominator. The result is the denominator becomes a difference of squares, one of which is the square of the radical.
[tex]\dfrac{4\sqrt{b}}{\sqrt{3}-b}=\dfrac{4\sqrt{b}}{\sqrt{3}-b}\cdot\dfrac{\sqrt{3}+b}{\sqrt{3}+b}=\boxed{\dfrac{4\sqrt{3b}+4b\sqrt{b}}{3-b^2}}[/tex]
