Question Number 66855 by John Kaloki Musau last updated on 20/Aug/19
$${simplify} \\ $$$$\frac{{p}^{\mathrm{2}} +\mathrm{2}{pq}+{q}^{\mathrm{2}} }{{p}^{\mathrm{3}} −{pq}^{\mathrm{2}} +{p}^{\mathrm{2}} {q}−{q}^{\mathrm{3}} } \\ $$
Answered by $@ty@m123 last updated on 20/Aug/19
$$\frac{\left({p}+{q}\right)^{\mathrm{2}} }{{p}^{\mathrm{3}} −{q}^{\mathrm{3}} +{pq}\left({p}−{q}\right)} \\ $$$$\frac{\left({p}+{q}\right)^{\mathrm{2}} }{\left({p}−{q}\right)\left({p}^{\mathrm{2}} +{pq}+{q}^{\mathrm{2}} \right)+{pq}\left({p}−{q}\right)} \\ $$$$\frac{\left({p}+{q}\right)^{\mathrm{2}} }{\left({p}−{q}\right)\left({p}^{\mathrm{2}} +{pq}+{q}^{\mathrm{2}} +{pq}\right)} \\ $$$$\frac{\left({p}+{q}\right)^{\mathrm{2}} }{\left({p}−{q}\right)\left({p}+{q}\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{{p}−{q}} \\ $$
Commented by John Kaloki Musau last updated on 20/Aug/19
$${wow}!\:{u}\:{got}\:{it}.{thanks} \\ $$
Answered by Mr Jor last updated on 25/Aug/19
$$\frac{\boldsymbol{{p}}^{\mathrm{2}} +\boldsymbol{{pq}}+\boldsymbol{{pq}}+\boldsymbol{{q}}^{\mathrm{2}} }{\boldsymbol{{p}}\left(\boldsymbol{{p}}^{\mathrm{2}} −\boldsymbol{{q}}^{\mathrm{2}} \right)+\boldsymbol{{q}}\left(\boldsymbol{{p}}^{\mathrm{2}} −\boldsymbol{{q}}^{\mathrm{2}} \right)} \\ $$$$\frac{\boldsymbol{{p}}\left(\boldsymbol{{p}}+\boldsymbol{{q}}\right)+\boldsymbol{{q}}\left(\boldsymbol{{p}}+\boldsymbol{{q}}\right)}{\left(\boldsymbol{{p}}+\boldsymbol{{q}}\right)\left(\boldsymbol{{p}}^{\mathrm{2}} −\boldsymbol{{q}}^{\mathrm{2}} \right)} \\ $$$$\frac{\left(\boldsymbol{{p}}+\boldsymbol{{q}}\right)\left(\boldsymbol{{p}}+\boldsymbol{{q}}\right)}{\left(\boldsymbol{{p}}+\boldsymbol{{q}}\right)\left(\boldsymbol{{p}}−\boldsymbol{{q}}\right)\left(\boldsymbol{{p}}+\boldsymbol{{q}}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\boldsymbol{{p}}−\boldsymbol{{q}}} \\ $$