Question 116

If $$\frac{a}{b}+\frac{b}{a}-1=0$$, then the value of $$a^{3}+b^{3}$$ is

Solution

$$\frac{a}{b} + \frac{b}{a} -1 = 0$$

$$\frac{a^{2}+b^{2}}{ab} -1 = 0$$

$$a^{2}+b^{2} - ab = 0$$

We know $$a^{3}+b^{3}={(a+b)}{(a^{2}+b^{2}}{-ab)} $$

As $$a^{2}+b^{2} - ab = 0$$, therefore $$a^{3}+b^{3}={(a+b)}{(a^{2}+b^{2}}{-ab)} =0$$

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