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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