Question Number 215504 by alephnull last updated on 08/Jan/25
$$\mathrm{simplify}\:{a}−\left\{{b}^{{c}−{b}} +\langle\left({b}^{{c}−{b}} \right)^{\mathrm{2}} +\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} \rangle+\frac{{a}}{\mathrm{2}}\right\}+{c}^{{ab}−{c}} −{c}^{{a}−{c}} \\ $$
Answered by MrGaster last updated on 08/Jan/25
$$\mathrm{simplify}\:{a}−\left\{{b}^{{c}−{b}} +\langle\left({b}^{{c}−{b}} \right)^{\mathrm{2}} +\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} \rangle+\frac{{a}}{\mathrm{2}}\right\}+{c}^{{ab}−{c}} −{c}^{{a}−{c}} \\ $$$${a}−{b}^{{c}−{a}} −\left({b}^{\mathrm{2}\left({c}−{a}\right)} +\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} \right)−\frac{{a}}{\mathrm{2}}+{c}^{{ab}−{c}} −{c}^{{a}−{c}} \\ $$$${a}−{b}^{{c}−{a}} −{b}^{\mathrm{2}\left({c}−{a}\right)} −\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−\frac{{a}}{\mathrm{2}}+{a}^{{ab}−{c}} −{c}^{{e}−{c}} \\ $$$$\left({a}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}−\frac{{a}}{\mathrm{2}}\right)−{b}^{{c}−{b}} −{b}^{\mathrm{2}\left({c}−{b}\right)} +{c}^{{ab}−{c}} −{c}^{{a}−{c}} \\ $$$$\left(\frac{\mathrm{2}{a}−{a}^{\mathrm{2}} }{\mathrm{4}}\right)−{b}^{{c}−{b}} −{b}^{\mathrm{2}\left({c}−{a}\right)} +{e}^{{ab}+{c}} −{c}^{{a}−{c}} \\ $$$$\frac{\mathrm{2}{a}−{a}^{\mathrm{2}} }{\mathrm{4}}−{b}^{{c}−{b}} −{b}^{\mathrm{2}\left({c}−{a}\right)} +{e}^{{ab}+{c}} −{c}^{{a}−{c}} \checkmark \\ $$
Commented by alephnull last updated on 09/Jan/25
$${thNks}\:{you} \\ $$$$ \\ $$