The value of \(\frac{(2.7)^3-(2.2)^3}{(2.7)^2+2.7\times2.2+(2.2)^2}\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.5
The expression has the form of the identity \(a^3-b^3=(a-b)(a^2+ab+b^2)\), with \(a=2.7\) and \(b=2.2\). The denominator is the second factor, so it cancels.
\(\displaystyle\frac{a^3-b^3}{a^2+ab+b^2}=\frac{(a-b)(a^2+ab+b^2)}{a^2+ab+b^2}=a-b\)
\(a-b=2.7-2.2=0.5\)
Tip: When the numerator is a difference (or sum) of cubes and the denominator looks like \(a^2\pm ab+b^2\), look for the identity before multiplying anything out; the answer is then just \(a-b\) (or \(a+b\)).