RRB 2009 Kolkata

Question 38

The value of $$\frac{(0.96)^{3}-(0.1)^{3}}{(0.96)^{2}+0.096+(0.1)^{2}}$$ is

Solution

$$a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$$

using this relation the given equation can be expanded as

$$\frac{(.96-.1)(.96^{2}+.096+.1^{2})}{.96^{2}+.096+.1^{2}}=.96 - .1=.86$$

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