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Answer:

a = 0 \: and \: b = 1

Step-by-step EXPLANATION:

LHS=\FRAC{7+3\sqrt{5}}{3+\sqrt{5}}-\frac{7-3\sqrt{5}}{3-\sqrt{5}}=\frac{(7+3\sqrt{5})(3-\sqrt{5})-(7-3\sqrt{5})(3+\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}=\frac{21-7\sqrt{5}+9\sqrt{5}-15-(21+7\sqrt{5}-9\sqrt{5}-15)}{3^{2}-(\sqrt{5})^{2}}

=\frac{21-7\sqrt{5}+9\sqrt{5}-15-21-7\sqrt{5}+9\sqrt{5}+15)}{9-5}

=\frac{4\sqrt{5}}{4}

=\sqrt{5}---(1)

RHS = a+\sqrt{5}b--(2)

Now\\0+\sqrt{5}=a+\sqrt{5}\\from\:(1)\:and\:(2)

a = 0 \: and \: b = 1

Therefore,

a = 0 \: and \: b = 1


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