Question 66

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

Solution

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

=> $$\frac{a^2+b^2}{ab}=1$$

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

We know that, $$(a^3+b^3)=(a+b)(a^2+b^2-ab)$$

Substituting value from equation (i), we get :

=> $$(a^3+b^3)=(a+b)(ab-ab)$$

=> $$(a^3+b^3)=(a+b) \times 0=0$$

=> Ans - (B)

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