Question Number 220007 by Noorzai last updated on 04/May/25
Commented by MrGaster last updated on 04/May/25
conditions are not enough and the meaning is unclear, please add or upload again if there is anything that has not been added.
Answered by efronzo1 last updated on 04/May/25
$$\:{a}^{\mathrm{3}} −{b}^{\mathrm{3}} \:=\:\left({a}−{b}\right)\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by Frix last updated on 04/May/25
$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{is}\:{r}^{\mathrm{2}} \pi \\ $$
Answered by Frix last updated on 04/May/25
$${a}^{\mathrm{2}} −{b}^{\mathrm{3}} =\left({a}−{b}^{\frac{\mathrm{3}}{\mathrm{2}}} \right)\left({a}+{b}^{\frac{\mathrm{3}}{\mathrm{2}}} \right) \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{3}} =\left({a}^{\frac{\mathrm{2}}{\mathrm{3}}} −{b}\right)\left({a}^{\frac{\mathrm{4}}{\mathrm{3}}} +{a}^{\frac{\mathrm{2}}{\mathrm{3}}} {b}+{b}^{\mathrm{2}} \right)= \\ $$$$=\left({a}^{\frac{\mathrm{2}}{\mathrm{3}}} −{b}\right)\left(\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){a}^{\frac{\mathrm{2}}{\mathrm{3}}} +{b}\right)\left(\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right){a}^{\frac{\mathrm{2}}{\mathrm{3}}} +{b}\right) \\ $$