Question : What is the value of $\frac{(1.1)^{3}+(0.7)^{3}}{(1.1)^{2}-1.1\times 0.7+(0.7)^{2}}?$
Option 1: 0.4
Option 2: 0.7
Option 3: 1.1
Option 4: 1.8
Correct Answer: 1.8
Solution :
Given: $\frac{(1.1)^{3}+(0.7)^{3}}{(1.1)^{2}-1.1\times 0.7+(0.7)^{2}}$
Using identity: $(a^3+b^3)=(a+b)(a^2-ab+b^2)$
$\frac{(1.1)^{3}+(0.7)^{3}}{(1.1)^{2}-1.1\times 0.7+(0.7)^{2}}$
= $\frac{(1.1+0.7)(1.1^2-1.1\times 0.7+0.7^2)}{(1.1)^{2}-1.1\times 0.7+(0.7)^{2}}$
= $(1.1+0.7)=1.8$
Hence, the correct answer is 1.8.
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