Question : If $a^3+b^3=9$ and $a+b=3$, then the value of $\frac{1}a+\frac{1}b$ is:
Option 1: $\frac{1}2$
Option 2: $\frac{3}2$
Option 3: $\frac{5}2$
Option 4: $–1$
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Correct Answer: $\frac{3}2$
Solution :
Given: That $a+b=3$ and $a^{3} + b^{3} = 9$
We know that $(a+ b)^{3}=a^3 + b^3 + 3ab(a + b)$
So, $a^3 + b^3 + 3ab(a + b) = 27$
⇒ $9 + 3ab(a + b) = 27$
⇒ $3ab(3) = 27 - 9$
⇒ $9ab=18$
⇒ $ab=2$
So, $\frac{1}a+\frac{1}b= \frac{(a+b)}{ab}$
= $\frac{3}{2}$
Hence, the correct answer is $\frac{3}{2}$.
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