Question

If $\frac{{37}^3-{28}^3}{{37}^2+37.28+{28}^2}=\_\_\_\_\_\_\_\_\_\_\_\_\_.$

Answer 1

$$\begin{gathered} \frac{{37}^3-{28}^3}{{37}^2+37.28+{28}^2} \\ =\frac{\left(37-28\right)\left({37}^2+{28}^2+\left(37\right)\left(28\right)\right)}{{37}^2+37.28+{28}^2}\left(\mathrm{Using}\mathrm{the}\mathrm{identity}:a^3-b^3=\left(a-b\right)\left(a^2+b^2+ab\right)\right) \\ =\frac{\left(37-28\right)}{1} \\ =9 \end{gathered}$$

Hence, if $\frac{{37}^3-{28}^3}{{37}^2+37.28+{28}^2}=\underline{9}.$