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Question Number 171243 by mathlove last updated on 11/Jun/22

Commented by infinityaction last updated on 11/Jun/22

$$6???$$

Commented by mathlove last updated on 11/Jun/22

$$how???$$

Commented by infinityaction last updated on 12/Jun/22

$$\frac{1}{a^3}+\frac{1}{b^3}=\frac{1}{a^3+b^3+c^3}-\frac{1}{c^3}$$$$\frac{a^3+b^3}{a^3{b}^3}=\frac{c^3-(a^3+b^3+c^3)}{c^3(a^3+b^3+c^3)}$$$$a^3+b^3+c^3=-\frac{a^3{b}^3}{c^3}$$$$similarly$$$$a^3+b^3+c^3=-\frac{b^3{c}^3}{a^3}$$$$and$$$$a^3+b^3+c^3=-\frac{c^3{a}^3}{b^3}$$$$so$$$$a=b=c$$$$\frac{a^{33}+a^{33}}{a^{33}}+\frac{a^{33}+a^{33}}{a^{33}}+\frac{a^{33}+a^{33}}{a^{33}}$$$$2+2+2=6$$$$pleasecheckmysolution$$

Answered by som(math1967) last updated on 11/Jun/22

$$\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{1}{p+q+r}[let{a}^3=p$$$$b^3=q,c^3=r]$$$$\Rightarrow (p+q+r)(pq+pr+qr)-pqr=0$$$$\Rightarrow pq(p+q)+pqr+p^{2}r+pqr+{pr}^2$$$$pqr+q^{2}r+{qr}^2-pqr=0$$$$\Rightarrow pq(p+q)+pr(p+q)+r^2(p+q)$$$$qr(p+q)=0$$$$\Rightarrow (p+q)(pq+pr+r^2+qr)=0$$$$\Rightarrow (p+q)(q+r)(p+r)=0$$$$ifp+q=0$$$$\therefore a^3+b^3=0$$$$a^3=-b^3$$$$a^{33}=-b^{33}\Rightarrow a^{33}+b^{33}=0$$$$\therefore \frac{{\mathit{a}}^{33}+{\mathit{b}}^{33}}{{\mathit{c}}^{33}}+\frac{{\mathit{b}}^{33}+{\mathit{c}}^{33}}{{\mathit{a}}^{33}}+\frac{{\mathit{c}}^{33}+{\mathit{a}}^{33}}{{\mathit{b}}^{33}}$$$$=0+\frac{{\mathit{b}}^{33}+{\mathit{c}}^{33}}{{\mathit{a}}^{33}}-\frac{{\mathit{c}}^{33}+{\mathit{a}}^{33}}{{\mathit{a}}^{33}}$$$$=\frac{{\mathit{b}}^{33}+{\mathit{c}}^{33}-{\mathit{c}}^{33}-{\mathit{a}}^{33}}{{\mathit{a}}^{33}}$$$$=\frac{-2a^{33}}{a^{33}}=-2\mathit{a}\mathit{n}\mathit{s}$$$$\mathit{i}\mathit{f}(\mathit{b}+\mathit{c})=0\mathit{o}\mathit{r}\mathit{c}+\mathit{a}=0$$$$\mathit{g}\mathit{i}\mathit{v}\mathit{e}\mathit{s}\mathit{s}\mathit{a}\mathit{m}\mathit{e}\mathit{r}\mathit{e}\mathit{s}\mathit{u}\mathit{l}\mathit{t}$$$$$$

Commented by infinityaction last updated on 11/Jun/22

$$whereiamwrongsir???$$

Commented by som(math1967) last updated on 11/Jun/22

$$\frac{-a^3{b}^3}{c^3}=\frac{-b^3{c}^3}{a^3}=\frac{-c^3{a}^3}{b^3}$$$$\Rightarrow \frac{-1}{c^6}=\frac{-1}{a^6}=\frac{-1}{b^6}$$$$\therefore a^6=b^6=c^6$$$$nowif{a}^6=b^6=c^6$$$$thenitmaynota=b=c$$

Commented by infinityaction last updated on 11/Jun/22

$$thankyousir$$

Commented by mathlove last updated on 12/Jun/22

$$thanks$$

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