Find the value of a and b, when a+b root 15= root 5+root 3upon root 5-root3
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

R.H.S,
On rationalizing the denominator we get,

On comparing L.H.S and R.H.S we get,

R.H.S,
On rationalizing the denominator we get,
On comparing L.H.S and R.H.S we get,
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