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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


Team Careers360 21st Jan, 2024
Answer (1)
Team Careers360 23rd Jan, 2024

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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