Question Number 208469 by hardmath last updated on 16/Jun/24
$$\mathrm{Find}:\:\:\:\:\:\frac{\mathrm{61}^{\mathrm{3}} \:\:+\:\:\mathrm{24}^{\mathrm{3}} }{\mathrm{61}^{\mathrm{3}} \:\:+\:\:\mathrm{37}^{\mathrm{3}} }\:\:=\:\:? \\ $$
Answered by Frix last updated on 16/Jun/24
$${a}=\mathrm{61};\:{b}=\mathrm{24};\:{a}−{b}=\mathrm{37} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\left({a}+{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right) \\ $$$${a}^{\mathrm{3}} +\left({a}−{b}\right)^{\mathrm{3}} = \\ $$$$=\left({a}+\left({a}−{b}\right)\right)\left({a}^{\mathrm{2}} −{a}\left({a}−{b}\right)+\left({a}−{b}\right)^{\mathrm{2}} \right)= \\ $$$$=\left(\mathrm{2}{a}−{b}\right)\left({a}^{\mathrm{2}} −{ab}+{b}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }{{a}^{\mathrm{3}} +\left({a}−{b}\right)^{\mathrm{3}} }=\frac{{a}+{b}}{\mathrm{2}{a}−{b}} \\ $$$$\frac{\mathrm{61}+\mathrm{24}}{\mathrm{2}×\mathrm{61}−\mathrm{24}}=\frac{\mathrm{85}}{\mathrm{98}} \\ $$
Commented by hardmath last updated on 17/Jun/24
$$\mathrm{thankyou}\:\mathrm{dear}\:\mathrm{professor} \\ $$