SSC CGL 2010 Tier 1 Question Paper - Shift 1

Question 118

If a = 11 and b = 9, then the value of $$\frac{a^2 + b^2 + ab}{a^3 - b^3}$$ is

Solution

$$\frac{a^2 + b^2 + ab}{a^3 - b^3} = \frac{a^2 + b^2 + ab}{(a - b)(a^2 + b^2 + ab)} = \frac{1}{a - b} = \frac{1}{11 - 9} = \frac{1}{2}$$

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