What is the value of \(\displaystyle\frac{(2.4)^{3}+(1.6)^{3}}{(2.4)^{2}-3.84+(1.6)^{2}}-\frac{(2.4)^{2}-(1.6)^{2}}{0.8}\)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 0
Take \(a=2.4\) and \(b=1.6\). The middle term 3.84 is \(a\times b\), so the first fraction is \(\frac{a^{3}+b^{3}}{a^{2}-ab+b^{2}}\), and 0.8 is \(a-b\), so the second is \(\frac{a^{2}-b^{2}}{a-b}\). Both reduce to \(a+b\).
\(ab=2.4\times1.6=3.84\) and \(a-b=2.4-1.6=0.8\)
\(\displaystyle\frac{a^{3}+b^{3}}{a^{2}-ab+b^{2}}=\frac{(a+b)(a^{2}-ab+b^{2})}{a^{2}-ab+b^{2}}=a+b=4\)
\(\displaystyle\frac{a^{2}-b^{2}}{a-b}=\frac{(a-b)(a+b)}{a-b}=a+b=4\)
\(4-4=0\)