Given, 3x - y = 3 ........ (I) On dividing equation (I) by '3', we have; {x - (y/3)} = 1 On cubing both sides, we have; x3 - (y3/27) - 3 X x X (y/3) X {x - (y/3)} = 1 Or, x3 - (y3/27) = 1 + xy X 1 Or, x3 - (y3/27) = 1 + (2/3) Or, x3 - (y3/27) = (5/3)
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