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Solution
$$x + y = 3$$
$$y=3-x$$ ---1
and,
$$xy = 2$$ ---2
So putting 1 in 2,
$$x(3-x) =2$$
$$x^2-3x+2=0$$
Solving this we get x=2 and -1 and taking positive values of x=2 in eq.1 we get $$y=1$$
So,
$$x^3 —y^3=(x-y)(x^2+xy+y^2)$$
$$=(2-1)(2^2+2\times1+1^2))$$
$$=7$$
Option D is correct.
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