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Question Number 60984 by naka3546 last updated on 28/May/19

$$\frac{a}{a-b}+\frac{b}{b-c}+\frac{c}{c-a}=4$$$${ab}^2+{bc}^2+abc+{ca}^2=a^{2}b+b^{2}c+c^{2}a$$$${\left(\frac{a}{a-b}\right)}^3+{\left(\frac{b}{b-c}\right)}^3+{\left(\frac{c}{c-a}\right)}^3=?$$$$$$

Commented by naka3546 last updated on 28/May/19

$$73?$$

Commented by Prithwish sen last updated on 28/May/19

$${\mathrm{ab}}^2+{\mathrm{bc}}^2+\mathrm{abc}+{\mathrm{ca}}^2={\mathrm{a}}^2\mathrm{b}+{\mathrm{b}}^2\mathrm{c}+{\mathrm{c}}^2\mathrm{a}$$$$\mathrm{ab}(\mathrm{a}-\mathrm{b})+\mathrm{bc}(\mathrm{b}-\mathrm{c})+\mathrm{ac}(\mathrm{c}-\mathrm{a})=\mathrm{abc}$$$$\frac{(\mathrm{a}-\mathrm{b})}{\mathrm{c}}+\frac{(\mathrm{b}-\mathrm{c})}{\mathrm{a}}+\frac{(\mathrm{c}-\mathrm{a})}{\mathrm{b}}=1$$$$\frac{-[(\mathrm{b}-\mathrm{c})+(\mathrm{c}-\mathrm{a})]}{\mathrm{c}}+\frac{(\mathrm{b}-\mathrm{c})}{\mathrm{a}}+\frac{(\mathrm{c}-\mathrm{a})}{\mathrm{b}}=1$$$$(\mathrm{b}-\mathrm{c})[\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{c}}]+(\mathrm{c}-\mathrm{a})[\frac{1}{\mathrm{b}}-\frac{1}{\mathrm{c}}]=1$$$$\frac{(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})}{\mathrm{c}}[\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}}]=1$$$$\frac{\mathrm{abc}}{(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})}=-1$$$$\mathrm{Now}\mathrm{let}$$$$\mathrm{A}=\frac{\mathrm{a}}{(\mathrm{a}-\mathrm{b})},\mathrm{B}=\frac{\mathrm{b}}{(\mathrm{b}-\mathrm{c})},\mathrm{C}=\frac{\mathrm{c}}{(\mathrm{c}-\mathrm{a})}$$$$\therefore \mathrm{A}+\mathrm{B}+\mathrm{C}=4,\mathrm{ABC}=-1$$$$\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{calculate}$$$${\mathrm{A}}^3+{\mathrm{B}}^3+{\mathrm{C}}^3$$$$\mathrm{please}\mathrm{help}.$$$$$$

Commented by naka3546 last updated on 28/May/19

$$\frac{a}{a-b}+\frac{b}{b-c}+\frac{c}{c-a}=4$$$$\Rightarrow 3+(\frac{b}{a-b}+\frac{c}{b-c}+\frac{a}{c-a})=4$$$$\iff (\frac{b}{a-b}+\frac{c}{b-c}+\frac{a}{c-a})=1$$$$\Rightarrow \frac{abc-(a^{2}b+b^{2}c+c^{2}a)}{({ab}^2+{bc}^2+{ca}^2)-(a^{2}b+b^{2}c+c^{2}a)}=1$$$$\iff abc={ab}^2+{bc}^2+{ca}^2...\left(i\right)$$$$$$$$abc+{ab}^2+{bc}^2+{ca}^2=a^{2}b+b^{2}c+c^{2}a$$$$\iff a^{2}b+b^{2}c+c^{2}a=2abc...\left(ii\right)$$$$$$$$\iff (a-b)(b-c)(c-a)=-abc...\left(iii\right)$$$$$$$${\left(\frac{a}{a-b}\right)}^3+{\left(\frac{b}{b-c}\right)}^3+{\left(\frac{c}{c-a}\right)}^3$$$$={(\frac{a}{a-b}+\frac{b}{b-c}+\frac{c}{c-a})}^3-3(\frac{a}{a-b}+\frac{b}{b-c}+\frac{c}{c-a})(\frac{ab}{(a-b)(b-c)}+\frac{bc}{(b-c)(c-a)}+\frac{ca}{(c-a)(a-b)})+3\left(\frac{abc}{(a-b)(b-c)(c-a)}\right)$$$$={(\frac{a}{a-b}+\frac{b}{b-c}+\frac{c}{c-a})}^3-3(\frac{a}{a-b}+\frac{b}{b-c}+\frac{c}{c-a})(-\frac{c-a}{c}-\frac{a-b}{a}-\frac{b-c}{b})+3\left(\frac{abc}{(a-b)(b-c)(c-a)}\right)$$$$={(\frac{a}{a-b}+\frac{b}{b-c}+\frac{c}{c-a})}^3+3(\frac{a}{a-b}+\frac{b}{b-c}+\frac{c}{c-a})(\frac{c-a}{c}+\frac{a-b}{a}+\frac{b-c}{b})+3\left(\frac{abc}{-abc}\right)$$$$={(\frac{a}{a-b}+\frac{b}{b-c}+\frac{c}{c-a})}^3+3(\frac{a}{a-b}+\frac{b}{b-c}+\frac{c}{c-a})(3-(\frac{a}{c}+\frac{b}{a}+\frac{c}{b}))+3\left(\frac{abc}{-abc}\right)$$$$={(\frac{a}{a-b}+\frac{b}{b-c}+\frac{c}{c-a})}^3+3(\frac{a}{a-b}+\frac{b}{b-c}+\frac{c}{c-a})(3-\left(\frac{a^{2}b+b^{2}c+c^{2}a}{abc}\right))-3$$$$={(\frac{a}{a-b}+\frac{b}{b-c}+\frac{c}{c-a})}^3+3(\frac{a}{a-b}+\frac{b}{b-c}+\frac{c}{c-a})(3-\left(\frac{2abc}{abc}\right))-3$$$$={\left(4\right)}^3+3\left(4\right)(3-2)-3$$$$=64+12-3$$$$=73$$

Commented by MJS last updated on 28/May/19

$$\mathrm{how}\mathrm{do}\mathrm{you}\mathrm{get}\left(i\right)?$$$$\frac{b}{a-b}+\frac{c}{b-c}+\frac{a}{c-a}\ne \frac{abc-(a^{2}b+b^{2}c+c^{2}a)}{({ab}^2+{bc}^2+{ca}^2)-(a^{2}b+b^{2}c+c^{2}a)}$$

Commented by Prithwish sen last updated on 28/May/19

$$\mathrm{great}\mathrm{sir}$$

Commented by naka3546 last updated on 28/May/19

$${ab}^2+{bc}^2+{ca}^2+abc=a^{2}b+b^{2}c+c^{2}a$$$$\Rightarrow (a+b+c)(ab+bc+ca)-(2abc+{ab}^2+{bc}^2+{ca}^2)=a^{2}b+b^{2}c+c^{2}a$$$$\iff (a+b+c)(ab+bc+ca)=2(a^{2}b+b^{2}c+c^{2}a)$$$$$$$$\frac{b}{a-b}+\frac{c}{b-c}+\frac{a}{c-a}=1$$$$\Rightarrow \frac{b(b-c)(c-a)+c(a-b)(c-a)+a(a-b)(b-c)}{(a-b)(b-c)(c-a)}=1$$$$\Rightarrow \frac{b^{2}c-{bc}^2-{ab}^2+abc+c^{2}a-{bc}^2-{ca}^2+abc+a^{2}b-{ab}^2-{ca}^2+abc}{(a-b)(b-c)(c-a)}=1$$$$\Rightarrow \frac{(a^{2}b+b^{2}c+c^{2}a)-2({ab}^2+{bc}^2+{ca}^2)+3abc}{(a-b)(b-c)(c-a)}=1$$$$\Rightarrow \frac{({ab}^2+{bc}^2+{ca}^2+abc)-2({ab}^2+{bc}^2+{ca}^2)+3abc}{(a-b)(b-c)(c-a)}=1$$$$\Rightarrow \frac{\left(4abc\right)-(a^{2}b+b^{2}c+c^{2}a)}{({ab}^2+{bc}^2+{ca}^2)-(a^{2}b+b^{2}c+c^{2}a)}=1$$$$4abc={ab}^2+{bc}^2+{ca}^2$$$$\iff a^{2}b+b^{2}c+c^{2}a=5abc$$$$Maybetherearemistakes.$$$$Typo,sir.$$

Commented by naka3546 last updated on 28/May/19

$${ab}^2+{bc}^2+{ca}^2+abc=a^{2}b+b^{2}c+c^{2}a$$$$\Rightarrow (a+b+c)(ab+bc+ca)-(2abc+{ab}^2+{bc}^2+{ca}^2)=a^{2}b+b^{2}c+c^{2}a$$$$\iff (a+b+c)(ab+bc+ca)=2(a^{2}b+b^{2}c+c^{2}a)$$$$$$$$\frac{b}{a-b}+\frac{c}{b-c}+\frac{a}{c-a}=1$$$$\Rightarrow \frac{b(b-c)(c-a)+c(a-b)(c-a)+a(a-b)(b-c)}{(a-b)(b-c)(c-a)}=1$$$$\Rightarrow \frac{b^{2}c-{bc}^2-{ab}^2+abc+c^{2}a-{bc}^2-{ca}^2+abc+a^{2}b-{ab}^2-{ca}^2+abc}{(a-b)(b-c)(c-a)}=1$$$$\Rightarrow \frac{(a^{2}b+b^{2}c+c^{2}a)-2({ab}^2+{bc}^2+{ca}^2)+3abc}{(a-b)(b-c)(c-a)}=1$$$$\Rightarrow \frac{({ab}^2+{bc}^2+{ca}^2+abc)-2({ab}^2+{bc}^2+{ca}^2)+3abc}{(a-b)(b-c)(c-a)}=1$$$$\Rightarrow \frac{\left(4abc\right)-(a^{2}b+b^{2}c+c^{2}a)}{({ab}^2+{bc}^2+{ca}^2)-(a^{2}b+b^{2}c+c^{2}a)}=1$$$$4abc={ab}^2+{bc}^2+{ca}^2$$$$\iff a^{2}b+b^{2}c+c^{2}a=5abc$$$$Maybetherearemistakes.$$$$Typo,sir.$$

Commented by naka3546 last updated on 28/May/19

$$37?$$

Commented by MJS last updated on 28/May/19

$${\mathrm{I}}^{\prime }\mathrm{ll}\mathrm{have}\mathrm{to}\mathrm{look}\mathrm{into}\mathrm{it}\mathrm{again}...$$

Answered by tanmay last updated on 28/May/19

$$\frac{a}{a-b}+\frac{b}{b-c}+\frac{c}{c-a}=4$$$$\frac{1}{1-\frac{b}{a}}+\frac{1}{1-\frac{c}{b}}+\frac{1}{1-\frac{a}{c}}=4$$$$u=\frac{b}{a}v=\frac{c}{b}w=\frac{a}{c}uvw=1$$$$\frac{1}{1-u}+\frac{1}{1-v}+\frac{1}{1-w}=4$$$$\frac{1}{1-u}-1+\frac{1}{1-v}-1+\frac{1}{1-w}-1=1$$$$\frac{u}{1-u}+\frac{v}{1-v}+\frac{w}{1-w}=1....\left(1\right)$$$${\left(\frac{1}{1-u}\right)}^3+{\left(\frac{1}{1-v}\right)}^3+{\left(\frac{1}{1-w}\right)}^3=???$$$${ab}^2+{bc}^2+{ca}^2+abc-a^{2}b-b^{2}c-c^{2}a=0$$$$ab(b-a)+bc(c-b)+ca(a-c)+abc=0$$$$a^2\times b(\frac{b}{a}-1)+b^2\times c(\frac{c}{b}-1)+c^{2}a(\frac{a}{c}-1)+abc=0$$$$a^3\times \frac{b}{a}(\frac{b}{a}-1)+b^3\times \frac{c}{b}(\frac{c}{b}-1)+c^3\times \frac{a}{c}(\frac{a}{c}-1)+abc=0$$$$a^{2}b(u-1)+b^{2}c(v-1)+c^{2}a(w-1)+abc=0$$$$wait...$$$$u=\frac{b}{a}v=\frac{c}{b}w=\frac{a}{c}uvw=1$$$$u:v:w=\frac{b}{a}:\frac{c}{b}:\frac{a}{c}=b^{2}c:{ac}^2:a^{2}b$$$$b^{2}c=uk{ac}^2=vk{a}^{2}b=wk$$$$a^3{b}^3{c}^3={uvwk}^3$$$$abc={\left(uvw\right)}^{\frac{1}{3}}k$$$$abc=k$$$$wk(u-1)+uk(v-1)+vk(w-1)+k=0$$$$uwk-wk+uvk-uk+vwk-vk+k=0$$

Commented by Prithwish sen last updated on 28/May/19

$$\mathrm{great}\mathrm{thinking}\mathrm{sir}.$$

Commented by tanmay last updated on 28/May/19

$$k(uw+uv+vw-u-v-w)=0$$$$k\ne 0$$$$uw+uv+vw-u-v-k=0$$$$u+v+k=uv+vw+uw$$$$wait...$$

Answered by MJS last updated on 29/May/19

$$\mathrm{let}b=pa\wedge c=qa$$$$\left(1\right)\Rightarrow (3-2q)p^2+(3q^2-3q-2)p+q(3-2q)=0$$$$\left(2\right)\Rightarrow (1-q)p^2+(q^2+q-1)p+q(1-q)=0$$$$\Rightarrow$$$$q=.198062p=\{\begin{array}{l}.307979\\ .643104\end{array}$$$$q=1.55496p=\{\begin{array}{l}.307979\\ 5.04892\end{array}$$$$q=3.24698p=\{\begin{array}{l}.643104\\ 5.04892\end{array}$$$$\mathrm{in}\mathrm{all}\mathrm{cases}\mathrm{we}\mathrm{get}\mathrm{the}\mathrm{result}25$$$$\mathrm{btw}\frac{1}{.198...}=5.04...;\frac{1}{.307...}=3.24...;\frac{1}{.643...}=1.55...$$

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