Question : Determine the value of $\small \left (\frac{1}{r}+\frac{1}{s} \right)$, when $r^{3}+s^{3}=0$ and $r+s=6$.
Option 1: 0
Option 2: 0.5
Option 3: 1
Option 4: 6
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Correct Answer: 0.5
Solution : We know that, $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$ So, we can write, $(r + s)^3 = r^3 + s^3 + 3rs(r + s)$ Substituting given values, we have, $6^3 = 0 + 3rs\times 6$ ⇒ $216 = 18rs$ ⇒ $rs = 12$ Now, $(\frac{1}r+ \frac{1}{s}) = \frac{(r + s)}{rs}$ ⇒ $(\frac{1}r+ \frac{1}{s}) = \frac{6}{12}$ = 0.5 Hence, the correct answer is 0.5.
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