If a=11 and b=9 , then the value of ≤ft((a^2+b^2+ab)/(a^3-b^3)) is:

The correct answer is Option (2) → \(\displaystyle\frac{1}{2}\)

Use the identity \(a^3-b^3=(a-b)(a^2+ab+b^2)\). The numerator is the second factor, so it cancels and the expression is just \(\frac{1}{a-b}\).

• \(\displaystyle\frac{a^2+b^2+ab}{(a-b)(a^2+ab+b^2)}=\frac{1}{a-b}\)

• \(\displaystyle\frac{1}{11-9}=\frac{1}{2}\)

Tip: Before putting numbers into a long expression, look for an identity that cancels; here it saves working out \(11^3\) and \(9^3\).