Question Number 50163 by Meritguide1234 last updated on 14/Dec/18
Answered by peter frank last updated on 14/Dec/18
$$\mathrm{let}\:\mathrm{x}=\frac{\mathrm{b}−\mathrm{c}}{\mathrm{a}}\:\:\:\: \\ $$$$\:\frac{\mathrm{1}}{\mathrm{x}}=\frac{\mathrm{a}}{\mathrm{b}−\mathrm{c}}\:\: \\ $$$$\mathrm{y}=\frac{\mathrm{c}−\mathrm{a}}{\mathrm{b}} \\ $$$$\frac{\mathrm{1}}{\mathrm{y}}=\frac{\mathrm{b}}{\mathrm{c}−\mathrm{a}} \\ $$$$\mathrm{z}=\frac{\mathrm{a}−\mathrm{b}}{\mathrm{c}} \\ $$$$\frac{\mathrm{1}}{\mathrm{z}}=\frac{\mathrm{c}}{\mathrm{a}−\mathrm{b}} \\ $$$$\left(\mathrm{x}+\mathrm{y}+\mathrm{z}\right)\left(\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{y}}+\frac{\mathrm{1}}{\mathrm{z}}\right) \\ $$$$\mathrm{3}+\frac{\mathrm{x}}{\mathrm{y}}+\frac{\mathrm{y}}{\mathrm{x}}+\frac{\mathrm{z}}{\mathrm{x}}+\frac{\mathrm{x}}{\mathrm{z}}+\frac{\mathrm{y}}{\mathrm{z}}+\frac{\mathrm{z}}{\mathrm{y}} \\ $$$$\mathrm{pls}\:{wait}… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18
![(p+q+r)((1/p)+(1/q)+(1/r)) =3+p((1/q)+(1/r))+q((1/r)+(1/p))+r((1/p)+(1/q)) =3+(p/q)+(r/q)+(p/r)+(q/r)+(q/p)+(r/p) =3+((p+r)/q)+((p+q)/r)+((q+r)/p) ((p+r)/q)=((((b−c)/a)+((a−b)/c))/((c−a)/b)) =((b(bc−c^2 +a^2 −ab))/(ac(c−a))) =(b/(ac))×((b(c−a)−(c+a)(c−a))/((c−a))) =(b/(ac))×(((c−a)(b−c−a))/((c−a))) =(b/(ac))×2b =((2b^3 )/(abc)) similarly others are ((2c^3 )/(abc)),((2a^3 )/(abc)) so 3+((2b^3 )/(abc))+((2c^3 )/(abc))+((2a^3 )/(abc)) =3+(2/(abc))(a^3 +b^3 +c^3 ) =3+(2/(abc))×3abc [when a+b+c=0 , a^3 +b^3 +c^3 =3abc] =3+6=9](https://www.tinkutara.com/question/Q50172.png)
$$\left({p}+{q}+{r}\right)\left(\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}\right) \\ $$$$=\mathrm{3}+{p}\left(\frac{\mathrm{1}}{{q}}+\frac{\mathrm{1}}{{r}}\right)+{q}\left(\frac{\mathrm{1}}{{r}}+\frac{\mathrm{1}}{{p}}\right)+{r}\left(\frac{\mathrm{1}}{{p}}+\frac{\mathrm{1}}{{q}}\right) \\ $$$$=\mathrm{3}+\frac{{p}}{{q}}+\frac{{r}}{{q}}+\frac{{p}}{{r}}+\frac{{q}}{{r}}+\frac{{q}}{{p}}+\frac{{r}}{{p}} \\ $$$$=\mathrm{3}+\frac{{p}+{r}}{{q}}+\frac{{p}+{q}}{{r}}+\frac{{q}+{r}}{{p}} \\ $$$$\frac{{p}+{r}}{{q}}=\frac{\frac{{b}−{c}}{{a}}+\frac{{a}−{b}}{{c}}}{\frac{{c}−{a}}{{b}}} \\ $$$$=\frac{{b}\left({bc}−{c}^{\mathrm{2}} +{a}^{\mathrm{2}} −{ab}\right)}{{ac}\left({c}−{a}\right)} \\ $$$$=\frac{{b}}{{ac}}×\frac{{b}\left({c}−{a}\right)−\left({c}+{a}\right)\left({c}−{a}\right)}{\left({c}−{a}\right)} \\ $$$$=\frac{{b}}{{ac}}×\frac{\left({c}−{a}\right)\left({b}−{c}−{a}\right)}{\left({c}−{a}\right)} \\ $$$$=\frac{{b}}{{ac}}×\mathrm{2}{b}\:\: \\ $$$$=\frac{\mathrm{2}{b}^{\mathrm{3}} }{{abc}} \\ $$$${similarly}\:{others}\:{are}\:\frac{\mathrm{2}{c}^{\mathrm{3}} }{{abc}},\frac{\mathrm{2}{a}^{\mathrm{3}} }{{abc}} \\ $$$${so} \\ $$$$\mathrm{3}+\frac{\mathrm{2}{b}^{\mathrm{3}} }{{abc}}+\frac{\mathrm{2}{c}^{\mathrm{3}} }{{abc}}+\frac{\mathrm{2}{a}^{\mathrm{3}} }{{abc}} \\ $$$$=\mathrm{3}+\frac{\mathrm{2}}{{abc}}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right) \\ $$$$=\mathrm{3}+\frac{\mathrm{2}}{{abc}}×\mathrm{3}{abc}\:\:\left[{when}\:{a}+{b}+{c}=\mathrm{0}\:,\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{3}{abc}\right] \\ $$$$=\mathrm{3}+\mathrm{6}=\mathrm{9} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by peter frank last updated on 14/Dec/18
$$\mathrm{nice}\:\mathrm{work}\:\mathrm{sir} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Dec/18
$${thank}\:{you}… \\ $$