If a = frac{sqrt{5} + 1}{sqrt{5} - 1} and b = | vedclass

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(B) First,rationalize $a$ and $b$:$a = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} 	imes \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{(\sqrt{5} + 1)^{2}}{5 - 1} =

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$\frac{4}{3}$

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(B) First,rationalize $a$ and $b$:$a = \frac{\sqrt{5} + 1}{\sqrt{5} - 1} 	imes \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{(\sqrt{5} + 1)^{2}}{5 - 1} = \frac{5 + 1 + 2\sqrt{5}}{4} = \frac{6 + 2\sqrt{5}}{4} = \frac{3 + \sqrt{5}}{2}$$b = \frac{\sqrt{5} - 1}{\sqrt{5} + 1} 	imes \frac{\sqrt{5} - 1}{\sqrt{5} - 1} = \frac{(\sqrt{5} - 1)^{2}}{5 - 1} = \frac{5 + 1 - 2\sqrt{5}}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}$Now,calculate $a + b$ and $ab$:$a + b = \frac{3 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2} = \frac{6}{2} = 3$$ab = \left(\frac{3 + \sqrt{5}}{2}ight) \left(\frac{3 - \sqrt{5}}{2}ight) = \frac{9 - 5}{4} = \frac{4}{4} = 1$We know that $a^{2} + b^{2} = (a + b)^{2} - 2ab = (3)^{2} - 2(1) = 9 - 2 = 7$Substitute these into the expression:$\frac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}} = \frac{(a^{2} + b^{2}) + ab}{(a^{2} + b^{2}) - ab} = \frac{7 + 1}{7 - 1} = \frac{8}{6} = \frac{4}{3}$

If $a = \frac{\sqrt{5} + 1}{\sqrt{5} - 1}$ and $b = \frac{\sqrt{5} - 1}{\sqrt{5} + 1}$,then the value of $\frac{a^{2} + ab + b^{2}}{a^{2} - ab + b^{2}}$ is

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