Question Number 171437 by solomonwells last updated on 15/Jun/22
$$\mathrm{make}\:\boldsymbol{\mathrm{r}}\:\:\:\mathrm{the}\:\mathrm{subject}\:\mathrm{of}\:\mathrm{the}\:\mathrm{formula} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{y}}=\left(\frac{\boldsymbol{\mathrm{pr}}}{\boldsymbol{\mathrm{m}}}\:\:−\:\frac{\boldsymbol{\mathrm{p}}^{\mathrm{3}} }{\mathrm{1}}\right)^{−\mathrm{3}/\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{find}\:\boldsymbol{\mathrm{r}}\:\mathrm{if}\:\mathrm{y}=−\mathrm{8}\:\:\:,\:\:\mathrm{m}=−\mathrm{1},\:\:\mathrm{p}=\mathrm{3} \\ $$
Answered by alephzero last updated on 15/Jun/22
$${y}\:=\:\left(\frac{{pr}}{{m}}−{p}^{\mathrm{3}} \right)^{−\mathrm{3}/\mathrm{2}} \\ $$$$\Rightarrow\:{y}^{−\mathrm{2}/\mathrm{3}} \:=\:\frac{{pr}}{{m}}−\frac{{mp}^{\mathrm{3}} }{{m}} \\ $$$$\Rightarrow\:{y}^{−\mathrm{2}/\mathrm{3}} \:=\:\frac{{pr}−{mp}^{\mathrm{3}} }{{m}} \\ $$$$\Rightarrow{my}^{−\mathrm{2}/\mathrm{3}} \:=\:{pr}−{mp}^{\mathrm{3}} \\ $$$$\Rightarrow\:{pr}\:=\:{my}^{−\mathrm{2}/\mathrm{3}} +{mp}^{\mathrm{3}} \\ $$$$\Rightarrow\:{r}\:=\:\frac{{my}^{−\mathrm{2}/\mathrm{3}} +{mp}^{\mathrm{3}} }{{p}} \\ $$
Commented by alephzero last updated on 17/Jun/22
$${y}\:=\:−\mathrm{8},\:{m}\:=\:−\mathrm{1},\:{p}\:=\:\mathrm{3} \\ $$$$\Rightarrow\:{r}\:=\:\frac{−\left(−\mathrm{8}\right)^{−\mathrm{2}/\mathrm{3}} −\mathrm{3}^{\mathrm{3}} }{\mathrm{3}}\:= \\ $$$$=\:−\frac{\sqrt[{\mathrm{3}}]{\left(−\mathrm{8}\right)^{\mathrm{2}} }−\mathrm{27}}{\mathrm{3}}\:=\:−\frac{\sqrt[{\mathrm{3}}]{\mathrm{64}}−\mathrm{27}}{\mathrm{3}}\:= \\ $$$$=\:−\frac{\mathrm{4}−\mathrm{27}}{\mathrm{3}}\:=\:\frac{\mathrm{27}−\mathrm{4}}{\mathrm{3}}\:=\:\frac{\mathrm{23}}{\mathrm{3}}\:=\:\mathrm{7}\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Answered by elviselloin719A last updated on 18/Jun/22
$${Ans}:\:\mathrm{7}.\mathrm{667} \\ $$