If $$x^3+y^3=36$$​ and $$x+y=6$$​, then $$\frac{1}{x}+\frac{1}{y}=$$​ ?

Correct option is D

Given:

​$$x^3 + y^3 = 36$$​​

x + y = 6

We are asked to find $$\frac{1}{x} + \frac{1}{y}.$$​​

Formula Used:

$$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$$

$$x^2 + y^2 = (x + y)^2 - 2xy$$​​

Solution:

$$x^3 + y^3 = (x + y)(x^2 - xy + y^2) \\ \ \\36 = 6(x^2 - xy + y^2) \\ \ \\x^2 - xy + y^2 = 6 \quad ...(1)$$​

$$x^2 + y^2 = (x + y)^2 - 2xy \\ \ \\x^2 + y^2 = 6^2 - 2xy = 36 - 2xy$$​

By equation (1), we have

(36 - 2xy) - xy = 6

36 - 3xy = 6

-3xy = -30

xy = 10

​$$\frac{1}{x} + \frac{1}{y} = \frac{x + y}{xy} = \frac{6}{10} = \frac35$$​​