If $a = 11$, and $b = 9$, then the value of $\left(\frac{a^2+b^2+ab}{a^3-b^3}\right)$ is:
Answer & explanation
Correct answer: option 2
We have,
$a = 11$, and $b = 9$
So, put them in, $\left(\frac{a^2+b^2+ab}{a^3-b^3}\right)$
$\left(\frac{11^2+9^2+11×9}{11^3-9^3}\right)$
= \(\frac{301}{1331 - 729}\) = \(\frac{301}{602}\) = \(\frac{1}{2}\)