Question Number 171243 by mathlove last updated on 11/Jun/22
Commented by infinityaction last updated on 11/Jun/22
$$\mathrm{6}??? \\ $$
Commented by mathlove last updated on 11/Jun/22
$${how}??? \\ $$
Commented by infinityaction last updated on 12/Jun/22
$$\frac{\mathrm{1}}{{a}^{\mathrm{3}} }+\frac{\mathrm{1}}{{b}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} }−\frac{\mathrm{1}}{{c}^{\mathrm{3}} } \\ $$$$\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} }{{a}^{\mathrm{3}} {b}^{\mathrm{3}} }\:=\:\frac{{c}^{\mathrm{3}} −\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)}{{c}^{\mathrm{3}} \left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \:=\:−\frac{{a}^{\mathrm{3}} {b}^{\mathrm{3}} }{{c}^{\mathrm{3}} } \\ $$$${similarly} \\ $$$$\:\:\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \:=\:−\frac{{b}^{\mathrm{3}} {c}^{\mathrm{3}} }{{a}^{\mathrm{3}} } \\ $$$${and} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \:=\:−\frac{{c}^{\mathrm{3}} {a}^{\mathrm{3}} }{{b}^{\mathrm{3}} } \\ $$$${so} \\ $$$${a}={b}={c} \\ $$$$\frac{{a}^{\mathrm{33}} +{a}^{\mathrm{33}} }{{a}^{\mathrm{33}} }+\frac{{a}^{\mathrm{33}} +{a}^{\mathrm{33}} }{{a}^{\mathrm{33}} }+\frac{{a}^{\mathrm{33}} +{a}^{\mathrm{33}} }{{a}^{\mathrm{33}} } \\ $$$$\mathrm{2}+\mathrm{2}+\mathrm{2}=\:\mathrm{6} \\ $$$$\:{please}\:{check}\:{my}\:{solution} \\ $$
Answered by som(math1967) last updated on 11/Jun/22
![(1/p) +(1/q) +(1/r)=(1/(p+q+r)) [let a^3 =p b^3 =q ,c^3 =r] ⇒(p+q+r)(pq+pr+qr)−pqr=0 ⇒pq(p+q) +pqr +p^2 r+pqr+pr^2 pqr+q^2 r+qr^2 −pqr=0 ⇒pq(p+q)+pr(p+q)+r^2 (p+q) qr(p+q)=0 ⇒(p+q)(pq+pr+r^2 +qr)=0 ⇒(p+q)(q+r)(p+r)=0 if p+q=0 ∴a^3 +b^3 =0 a^3 =−b^3 a^(33) =−b^(33) ⇒a^(33) +b^(33) =0 ∴((a^(33) +b^(33) )/c^(33) ) +((b^(33) +c^(33) )/a^(33) ) +((c^(33) +a^(33) )/b^(33) ) =0+((b^(33) +c^(33) )/a^(33) ) −((c^(33) +a^(33) )/a^(33) ) = ((b^(33) +c^(33) −c^(33) −a^(33) )/a^(33) ) =((−2a^(33) )/a^(33) )=−2 ans if (b+c)=0 or c+a=0 gives same result](https://www.tinkutara.com/question/Q171277.png)
$$\:\frac{\mathrm{1}}{{p}}\:+\frac{\mathrm{1}}{{q}}\:+\frac{\mathrm{1}}{{r}}=\frac{\mathrm{1}}{{p}+{q}+{r}}\:\left[{let}\:{a}^{\mathrm{3}} ={p}\right. \\ $$$$\left.{b}^{\mathrm{3}} ={q}\:\:\:,{c}^{\mathrm{3}} ={r}\right] \\ $$$$\Rightarrow\left({p}+{q}+{r}\right)\left({pq}+{pr}+{qr}\right)−{pqr}=\mathrm{0} \\ $$$$\Rightarrow{pq}\left({p}+{q}\right)\:+{pqr}\:+{p}^{\mathrm{2}} {r}+{pqr}+{pr}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:{pqr}+{q}^{\mathrm{2}} {r}+{qr}^{\mathrm{2}} −{pqr}=\mathrm{0} \\ $$$$\Rightarrow{pq}\left({p}+{q}\right)+{pr}\left({p}+{q}\right)+{r}^{\mathrm{2}} \left({p}+{q}\right) \\ $$$$\:{qr}\left({p}+{q}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({p}+{q}\right)\left({pq}+{pr}+{r}^{\mathrm{2}} +{qr}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({p}+{q}\right)\left({q}+{r}\right)\left({p}+{r}\right)=\mathrm{0} \\ $$$${if}\:{p}+{q}=\mathrm{0} \\ $$$$\therefore{a}^{\mathrm{3}} +{b}^{\mathrm{3}} =\mathrm{0} \\ $$$${a}^{\mathrm{3}} =−{b}^{\mathrm{3}} \\ $$$${a}^{\mathrm{33}} =−{b}^{\mathrm{33}} \Rightarrow{a}^{\mathrm{33}} +{b}^{\mathrm{33}} =\mathrm{0} \\ $$$$\therefore\frac{\boldsymbol{{a}}^{\mathrm{33}} +\boldsymbol{{b}}^{\mathrm{33}} }{\boldsymbol{{c}}^{\mathrm{33}} }\:+\frac{\boldsymbol{{b}}^{\mathrm{33}} +\boldsymbol{{c}}^{\mathrm{33}} }{\boldsymbol{{a}}^{\mathrm{33}} }\:+\frac{\boldsymbol{{c}}^{\mathrm{33}} +\boldsymbol{{a}}^{\mathrm{33}} }{\boldsymbol{{b}}^{\mathrm{33}} } \\ $$$$=\mathrm{0}+\frac{\boldsymbol{{b}}^{\mathrm{33}} +\boldsymbol{{c}}^{\mathrm{33}} }{\boldsymbol{{a}}^{\mathrm{33}} }\:−\frac{\boldsymbol{{c}}^{\mathrm{33}} +\boldsymbol{{a}}^{\mathrm{33}} }{\boldsymbol{{a}}^{\mathrm{33}} } \\ $$$$=\:\frac{\boldsymbol{{b}}^{\mathrm{33}} +\boldsymbol{{c}}^{\mathrm{33}} −\boldsymbol{{c}}^{\mathrm{33}} −\boldsymbol{{a}}^{\mathrm{33}} }{\boldsymbol{{a}}^{\mathrm{33}} } \\ $$$$=\frac{−\mathrm{2}{a}^{\mathrm{33}} }{{a}^{\mathrm{33}} }=−\mathrm{2}\:\boldsymbol{{ans}} \\ $$$$\boldsymbol{{if}}\:\left(\boldsymbol{{b}}+\boldsymbol{{c}}\right)=\mathrm{0}\:\boldsymbol{{or}}\:\boldsymbol{{c}}+\boldsymbol{{a}}=\mathrm{0}\: \\ $$$$\boldsymbol{{gives}}\:\boldsymbol{{same}}\:\boldsymbol{{result}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by infinityaction last updated on 11/Jun/22
$${where}\:{i}\:{am}\:{wrong}\:{sir}\:??? \\ $$
Commented by som(math1967) last updated on 11/Jun/22
$$\:\frac{−{a}^{\mathrm{3}} {b}^{\mathrm{3}} }{{c}^{\mathrm{3}} }=\frac{−{b}^{\mathrm{3}} {c}^{\mathrm{3}} }{{a}^{\mathrm{3}} }=\frac{−{c}^{\mathrm{3}} {a}^{\mathrm{3}} }{{b}^{\mathrm{3}} } \\ $$$$\Rightarrow\frac{−\mathrm{1}}{{c}^{\mathrm{6}} }=\frac{−\mathrm{1}}{{a}^{\mathrm{6}} }=\frac{−\mathrm{1}}{{b}^{\mathrm{6}} } \\ $$$$\therefore\:{a}^{\mathrm{6}} ={b}^{\mathrm{6}} ={c}^{\mathrm{6}} \\ $$$${now}\:{if}\:{a}^{\mathrm{6}} ={b}^{\mathrm{6}} ={c}^{\mathrm{6}} \\ $$$${then}\:{it}\:{may}\:{not}\:{a}={b}={c} \\ $$
Commented by infinityaction last updated on 11/Jun/22
$${thank}\:{you}\:{sir} \\ $$
Commented by mathlove last updated on 12/Jun/22
$${thanks} \\ $$