Question Number 185529 by Rupesh123 last updated on 23/Jan/23
Answered by Frix last updated on 23/Jan/23
$${x}=\frac{\mathrm{1}}{{a}}\wedge{y}=\frac{\mathrm{1}}{{b}}\wedge{z}=\frac{\mathrm{1}}{{c}}\:\Rightarrow \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \\ $$$$\mathrm{2}\left({xy}+{yz}+{zx}\right)=\mathrm{0}\:\Rightarrow \\ $$$${z}=−\frac{{xy}}{{x}+{y}}\:\Leftrightarrow\:{c}=−\left({a}+{b}\right)\:\Rightarrow \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =−\mathrm{3}{ab}\left({a}+{b}\right)=\mathrm{3}{abc} \\ $$
Answered by som(math1967) last updated on 23/Jan/23
![(1/a^2 )+(1/b^2 )+(1/c^2 ) +2((1/(ab))+(1/(bc))+(1/(ca)))=(1/a^2 )+(1/b^2 )+(1/c^2 ) ⇒2(((a+b+c)/(abc)))=0 ⇒a+b+c=0 ⇒(a+b)=−c ⇒(a+b)^3 =−c^3 ⇒a^3 +b^3 +3ab(a+b)=−c^3 ⇒a^3 +b^3 +c^3 =3abc [∵a+b=−c] ∴a^3 +b^3 +c^3 is multiple of 3](https://www.tinkutara.com/question/Q185534.png)
$$\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\:+\mathrm{2}\left(\frac{\mathrm{1}}{{ab}}+\frac{\mathrm{1}}{{bc}}+\frac{\mathrm{1}}{{ca}}\right)=\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{c}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{2}\left(\frac{{a}+{b}+{c}}{{abc}}\right)=\mathrm{0} \\ $$$$\Rightarrow{a}+{b}+{c}=\mathrm{0} \\ $$$$\Rightarrow\left({a}+{b}\right)=−{c} \\ $$$$\Rightarrow\left({a}+{b}\right)^{\mathrm{3}} =−{c}^{\mathrm{3}} \\ $$$$\Rightarrow{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\mathrm{3}{ab}\left({a}+{b}\right)=−{c}^{\mathrm{3}} \\ $$$$\Rightarrow{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{3}{abc}\:\:\:\:\left[\because{a}+{b}=−{c}\right] \\ $$$$\therefore{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \:{is}\:{multiple}\:{of}\:\mathrm{3} \\ $$$$ \\ $$