Question Number 191546 by MATHEMATICSAM last updated on 25/Apr/23
$$\mathrm{If}\:{m}\:+\:\mathrm{1}\:=\:\sqrt{{n}}\:+\:\mathrm{3}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{m}^{\mathrm{3}} \:−\:\mathrm{6}{m}^{\mathrm{2}} \:+\:\mathrm{12}{m}\:−\mathrm{8}}{\:\sqrt{{n}}}\:−\:{n}\right) \\ $$
Answered by som(math1967) last updated on 25/Apr/23
![(1/2)×{(((m−2)^3 )/( (√n))) −n} (1/2)×{((((√n))^3 )/( (√n))) −n} [m+1=(√n)+3⇒m−2=(√n)] (1/2)×(n−n)=0](https://www.tinkutara.com/question/Q191547.png)
$$\frac{\mathrm{1}}{\mathrm{2}}×\left\{\frac{\left({m}−\mathrm{2}\right)^{\mathrm{3}} }{\:\sqrt{{n}}}\:−{n}\right\} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\left\{\frac{\left(\sqrt{{n}}\right)^{\mathrm{3}} }{\:\sqrt{{n}}}\:−{n}\right\}\:\left[{m}+\mathrm{1}=\sqrt{{n}}+\mathrm{3}\Rightarrow{m}−\mathrm{2}=\sqrt{{n}}\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\left({n}−{n}\right)=\mathrm{0} \\ $$$$ \\ $$