If the sum of squares of two real numbers is 41 and their sum is 9. Then the sum of cubes of these two numbers is
x + y = 9
$$x^2 + y^2$$ = 41
$$(x + y)^2 =Â x^2 + y^2 + 2xy$$
81 = 41 + 2xy
xy =20
x - y = $$\sqrt{(x + y)^2 - 4xy}$$ = 1
So, x=5 , y=4
$$x^3 + y^3$$ =Â $$5^3 + 4^3$$ = 189
So, the answer would be option c)189.
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