Question 148

If $$\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$$, then the value of $$a^{3}-b^{3}$$ is

Solution

Given : $$\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$$

=> $$\frac{1}{a+b}=\frac{a+b}{ab}$$

=> $$(a+b)^2=ab$$

=> $$(a^2+b^2+2ab)-ab=0$$

=> $$a^2+b^2+ab=0$$ -------------(i)

To find : $$a^{3}-b^{3}$$

= $$(a-b)(a^2+b^2+ab)$$

Using equation (i), we get :

=> $$(a-b)\times0=0$$

=> Ans - (C)

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