Question 68

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

Solution

x + y = 9

$$x^2 + y^2$$ = 41

$$(x + y)^2 = x^2 + y^2 + 2xy$$

81 = 41 + 2xy

xy =20

x - y = $$\sqrt{(x + y)^2 - 4xy}$$ = 1

So, x=5 , y=4

$$x^3 + y^3$$ = $$5^3 + 4^3$$ = 189

So, the answer would be option c)189.

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