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(3+√2/3-√2)-(3-√2/3+√2) = a+12√2/7b
=> [3+√2/3-√2×3+√2/3+√2]-[3-√2/3+√2×3-√2/3-
√2]
=> [(3+√2)²/(3)²-(√2)²]-[(3-√2)²/(3)²-(√2)²]
=> [9+2+6√2/9-2]-[9+2-6√2/9-2]
=> [11+6√2/7]-[11-6√2/7]
=> 11+6√2-(11-6√2)/7
=> 11+6√2-11+6√2/7
=> 2(6√2)/7
=> 12√2/7 = L.H.S.
Hence, a = 0 and b = 1.
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