Question : If $a=26$ and $b=22$, then the value of $\frac{a^3-b^3}{a^2-b^2}-\frac{3 a b}{a+b}$ is_______.
Option 1: $\frac{5}{3}$
Option 2: $\frac{13}{11}$
Option 3: $\frac{1}{3}$
Option 4: $\frac{11}{13}$
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Correct Answer: $\frac{1}{3}$
Solution :
Given, $a=26$ and $b=22$
Now, $\frac{a^3-b^3}{a^2-b^2}-\frac{3ab}{a+b}$
= $\frac{(a-b)(a^2+b^2-ab)}{(a-b)(a+b)}-\frac{3ab}{a+b}$
= $\frac{a^2+b^2-2ab}{a+b}$
= $\frac{(a-b)^2}{a+b}$
= $\frac{(26-22)^2}{(26+22)}$
= $\frac{4^2}{48}$
= $\frac{1}{3}$
Hence, the correct answer is $\frac{1}{3}$.
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