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Question Number 70040 by Shamim last updated on 30/Sep/19

$$\mathrm{If},{\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2(\frac{1}{{\mathrm{a}}^3}+\frac{1}{{\mathrm{b}}^3}+\frac{1}{{\mathrm{c}}^3})={\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3\mathrm{than}$$$$\mathrm{prove}\mathrm{that}\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{Successive}\mathrm{Proportional}.$$

Commented by Prithwish sen last updated on 01/Oct/19

$$(\frac{{\mathrm{b}}^2{\mathrm{c}}^2}{\mathrm{a}}-{\mathrm{a}}^3)+(\frac{{\mathrm{a}}^2{\mathrm{c}}^2}{\mathrm{b}}-{\mathrm{b}}^3)+(\frac{{\mathrm{a}}^2{\mathrm{b}}^2}{\mathrm{c}}-{\mathrm{c}}^3)=0$$$$\frac{{\mathrm{b}}^2{\mathrm{c}}^2-{\mathrm{a}}^4}{\mathrm{a}}+\frac{{\mathrm{a}}^2{\mathrm{b}}^2-{\mathrm{c}}^4}{\mathrm{c}}+\left(\frac{{\mathrm{a}}^2{\mathrm{c}}^2-{\mathrm{b}}^4}{\mathrm{c}}\right)=0$$$$\frac{{\mathrm{b}}^2{\mathrm{c}}^3-{\mathrm{a}}^4\mathrm{c}+{\mathrm{a}}^3{\mathrm{b}}^2-{\mathrm{ac}}^4}{\mathrm{ac}}+\frac{{\mathrm{a}}^2{\mathrm{c}}^2-{\mathrm{b}}^4}{\mathrm{c}}=0$$$$\frac{({\mathbf{b}}^2-\mathbf{ac})({\mathrm{a}}^3+{\mathrm{c}}^3)}{\mathrm{ac}}-\frac{({\mathbf{b}}^2-\mathbf{ac})({\mathrm{b}}^2+\mathrm{ac})}{\mathrm{c}}=0$$$$({\mathrm{b}}^2-\mathrm{ac})(\frac{{\mathrm{a}}^3+{\mathrm{c}}^3}{\mathrm{ac}}-\frac{{\mathrm{b}}^2+\mathrm{ac}}{\mathrm{c}})=0$$$$\mathbf{considering}\mathbf{only}$$$${\mathbf{b}}^2-\mathbf{ac}=0$$$$\Rightarrow \mathbf{a},\mathbf{b},\mathbf{c}\mathbf{are}\mathbf{in}\mathbf{proportion}.\mathbf{proved}$$

Commented by Shamim last updated on 01/Oct/19

$$\mathrm{tnks}$$

Commented by MJS last updated on 01/Oct/19

$$\mathrm{of}\mathrm{course}\mathrm{this}\mathrm{is}\mathrm{right}$$$$$$$$\mathrm{I}\mathrm{saw}\mathrm{that}\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{can}\mathrm{be}$$$$\mathrm{transformed}\mathrm{to}$$$$a^{4}bc+{ab}^{4}c+{abc}^4-a^3{b}^3-a^3{c}^3-b^3{c}^3=0\iff$$$$\iff (a^2-bc)(b^2-ac)(c^2-ab)=0$$$$\mathrm{but}\mathrm{I}\mathrm{thought}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{easy}\mathrm{to}\mathrm{factorize}$$$$$$$$\mathrm{on}\mathrm{the}\mathrm{other}\mathrm{hand}\mathrm{by}\mathrm{taking}\mathrm{the}\mathrm{result}\mathrm{and}$$$$\mathrm{simply}\mathrm{putting}b=pa,c=p^{2}a\mathrm{or}\mathrm{just}c=\frac{b^2}{a}\mathrm{we}$$$$\mathrm{easily}\mathrm{find}\mathrm{the}\mathrm{given}\mathrm{equation}\mathrm{is}\mathrm{true}$$$$$$$$\mathrm{but}\mathrm{I}\mathrm{thought}\mathrm{showing}\mathrm{that}\mathrm{with}b=pa\mathrm{and}$$$$c=qa\Rightarrow q=p^2\mathrm{is}\mathrm{more}\mathrm{convincing}$$

Answered by MJS last updated on 30/Sep/19

$$\mathrm{for}a,b,c\in \mathbb{R}\mathrm{we}\mathrm{can}\mathrm{find}p,q\in \mathbb{R}\Rightarrow b=pa\wedge c=qa$$$$$$$$a^2{b}^2{c}^2(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3})=a^3+b^3+c^3$$$$a^3{b}^3+a^3{c}^3+b^3{c}^3=a^{4}bc+{ab}^{4}c+{abc}^4$$$$a^{4}bc-a^3(b^3-c^3)+a(b^{4}c+{bc}^4)-b^3{c}^3=0$$$$b=pa\wedge c=qa,\mathrm{let}a⩽b⩽c\Rightarrow 1⩽p⩽q$$$$a^6(-p^3{q}^3+p^{4}q+{pq}^4-p^3-q^3+pq)=0$$$$a\ne 0$$$$-p^3{q}^3+p^{4}q+{pq}^4-p^3-q^3+pq=0$$$$(p^2-q)(p-q^2)(pq-1)=0$$$$\Rightarrow q=p^2[\vee q=\pm \sqrt{p}\vee q=\frac{1}{p}]$$$$\Rightarrow ⟨a,b,c⟩=⟨a,pa,p^{2}a⟩$$

Commented by Shamim last updated on 30/Sep/19

$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{very}\mathrm{hard}.\mathrm{others}\mathrm{rule}???$$

Commented by MJS last updated on 30/Sep/19

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{that}\mathrm{hard}.\mathrm{write}$$$$-p^3{q}^3+p^{4}q+{pq}^4-p^3-q^3+pq=0$$$$\mathrm{as}$$$${pq}^4-(p^3+1)q^3+p(p^3+1)q-p^3=0$$$$\mathrm{divide}\mathrm{by}p$$$$q^4-\frac{p^3+1}{p}{q}^3+(p^3+1)q-p^2=0$$$$\mathrm{we}\mathrm{know}\mathrm{if}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{an}\mathrm{easy}\mathrm{solution}{q}_1\Rightarrow$$$$q_1\mid p^2\Rightarrow \mathrm{we}\mathrm{have}\mathrm{to}\mathrm{try}\pm 1,\pm p,\pm p^2$$$$\Rightarrow q_1=p^2\mathrm{and}\mathrm{this}\mathrm{is}\mathrm{enough}\mathrm{because}\mathrm{we}$$$$\mathrm{set}a⩽b⩽c\Rightarrow 1⩽p⩽q\Rightarrow q\ne \pm \sqrt{p}\wedge q\ne \frac{1}{p}$$$$$$$$\mathrm{we}\mathrm{then}\mathrm{have}$$$$(q-p^2)(q^3-\frac{1}{p}{q}^2-pq+1)=0$$$$\mathrm{transform}\mathrm{to}$$$$(q-p^2)(q^2(q-\frac{1}{p})-p(q-\frac{1}{p}))=0$$$$\Rightarrow q_2=\frac{1}{p}$$$$(q-p^2)(q-\frac{1}{p})(q^2-p)=0$$$$\Rightarrow q_{3,4}=\pm \sqrt{p}$$

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