Question : If $x= 19$ and $y= 18$, then the value of $\frac{x^{2}+y^{2}+xy}{x^{3}-y^{3}}$ is:
Option 1: 1
Option 2: 37
Option 3: 324
Option 4: 361
Answer (1)
12th Jan, 2024
Correct Answer: 1
Solution :
Given: $x = 19$ and $y= 18$
Since, $x^{3}-y^{3} = (x-y)(x^{2}+y^{2}+xy)$
$\therefore\frac{x^{2}+y^{2}+xy}{x^{3}-y^{3}}=\frac{x^{2}+y^{2}+xy}{ (x-y)(x^{2}+y^{2}+xy)}=\frac{1}{x-y}=\frac{1}{19-18}= 1$
Hence, the correct answer is 1.
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