If x+y=3 and xy=2,then find the value of x^3- | vedclass

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(D) Given: $x+y=3$ and $xy=2$.We know that $(x-y)^2 = (x+y)^2 - 4xy$.Substituting the values: $(x-y)^2 = 3^2 - 4(2) = 9 - 8 = 1$.Thus,$x-y = 1$ (assum

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$7$

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(D) Given: $x+y=3$ and $xy=2$.We know that $(x-y)^2 = (x+y)^2 - 4xy$.Substituting the values: $(x-y)^2 = 3^2 - 4(2) = 9 - 8 = 1$.Thus,$x-y = 1$ (assuming $x > y$).Now,we use the identity $x^3-y^3 = (x-y)(x^2+y^2+xy)$.We can rewrite this as $x^3-y^3 = (x-y)((x+y)^2 - xy)$.Substituting the known values: $x^3-y^3 = 1 	imes (3^2 - 2) = 1 	imes (9 - 2) = 7$.

If $x+y=3$ and $xy=2$,then find the value of $x^3-y^3$.

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