If \(a=22\) and \(b=18\), then the value of \(\frac{a^2+b^2+ab}{a^3-b^3}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\frac{1}{4}\)
The denominator factorises by the identity \(a^3-b^3=(a-b)(a^2+ab+b^2)\), and its second factor is the numerator itself (the same three terms in another order). So the common factor cancels and the fraction reduces to \(\frac{1}{a-b}\):
Factorising the denominator: \(\displaystyle \frac{a^2+b^2+ab}{a^3-b^3}=\frac{a^2+ab+b^2}{(a-b)(a^2+ab+b^2)}=\frac{1}{a-b}\)
Putting in \(a=22\) and \(b=18\): \(\displaystyle \frac{1}{22-18}=\frac{1}{4}\)
Tip: When an expression has \(a^3-b^3\) (or \(a^3+b^3\)) below and \(a^2+ab+b^2\) (or \(a^2-ab+b^2\)) above, look for the identity first; it saves cubing 22 and 18.