Question 9
If $$a + b = 2$$, $$\frac{1}{a} + \frac{1}{b} = 18$$, then $$a^3 + b^3$$ lies between
Solution
Since $$\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab} = 18$$, we get $$ab = \frac{2}{18} = \frac{1}{9}$$. Then $$a^3 + b^3 = (a+b)^3 - 3ab(a+b) = 8 - 3 \times \frac{1}{9} \times 2 = 8 - \frac{2}{3} = \frac{22}{3}$$. As $$\frac{22}{3} \approx 7.33$$, the value lies between 7 and 8.
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