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Question Number 205527 by MATHEMATICSAM last updated on 23/Mar/24

$$\mathrm{If}\mathrm{the}\mathrm{roots}\mathrm{of}{ax}^2+bx+c=0\mathrm{are}\mathrm{one}$$$${\mathrm{another}}^{\prime }\mathrm{s}\mathrm{cube}\mathrm{then}\mathrm{show}\mathrm{that}$$$${(b^2-2ac)}^2=ac{(a+c)}^2.$$

Answered by A5T last updated on 23/Mar/24

$$Lettherootsbe\alpha and\beta .Then,\alpha ={\beta }^{3}and\beta ={\alpha }^3$$$$\alpha +\beta =\alpha +{\alpha }^3=\frac{-b}{a};\alpha \beta ={\alpha }^4=\frac{c}{a}\Rightarrow \alpha =\underset{-}{+}\sqrt[4]{\frac{c}{a}}$$$$\Rightarrow \underset{-}{+}\sqrt[4]{\frac{c}{a}}\underset{-}{+}\sqrt[4]{\frac{c^3}{a^3}}=\frac{-b}{a}\Rightarrow \sqrt{\frac{c}{a}}+\sqrt{\frac{c^3}{a^3}}+\frac{2c}{a}=\frac{b^2}{a^2}$$$$\Rightarrow \frac{c}{a}+\frac{c^3}{a^3}+\frac{2c^2}{a^2}=\frac{{(b^2-2ac)}^2}{a^4}$$$$\Rightarrow a^{3}c+{ac}^3+2a^2{c}^2=ac{(a+c)}^2={(b^2-2ac)}^2$$

Answered by Rasheed.Sindhi last updated on 24/Mar/24

$$\mathrm{AnOther}\mathrm{Way}$$$${(b^2-2ac)}^2=ac{(a+c)}^2$$$$\iff {\left(a^2(\frac{b^2}{a^2}-\frac{2ac}{a^2})\right)}^2=ac{\left(a(\frac{a}{a}+\frac{c}{a})\right)}^2$$$$\iff a^4{({(-\frac{b}{a})}^2-2\left(\frac{c}{a}\right))}^2=a^{3}c{(1+\frac{c}{a})}^2$$$$\iff {({(-\frac{b}{a})}^2-2\left(\frac{c}{a}\right))}^2=\frac{a^{3}c}{a^4}{(1+\frac{c}{a})}^2$$$$\iff {({(-\frac{b}{a})}^2-2\left(\frac{c}{a}\right))}^2=\frac{c}{a}{(1+\frac{c}{a})}^2$$$$\iff {({(\alpha +\beta )}^2-2\left(\alpha \beta \right))}^2=\alpha \beta {(1+\alpha \beta )}^2$$$$\iff {({\alpha }^2+{\beta }^2)}^2=\alpha \beta (1+2\alpha \beta +{\alpha }^2{\beta }^2)$$$$\iff {\alpha }^4+2{\alpha }^2{\beta }^2+{\beta }^4=\alpha \beta +2{\alpha }^2{\beta }^2+{\alpha }^3{\beta }^3$$$${\alpha }^3=\beta ,{\beta }^3=\alpha \Rightarrow {\alpha }^3{\beta }^3=\alpha \beta$$$${\alpha }^4=\alpha \cdot {\alpha }^3=\alpha \beta ,{\beta }^4={\beta }^3\cdot \beta =\alpha \beta$$$$\iff \alpha \beta +2{\alpha }^2{\beta }^2+\alpha \beta =\alpha \beta +2{\alpha }^2{\beta }^2+\alpha \beta$$$$\iff 2\alpha \beta +2{\alpha }^2{\beta }^2=2\alpha \beta +2{\alpha }^2{\beta }^2$$$$\mathcal{H}enceproved$$

Answered by Rasheed.Sindhi last updated on 24/Mar/24

$$\mathrm{Still}\mathrm{another}\mathrm{way}$$$$\alpha +\beta =\alpha +{\alpha }^3=-\frac{b}{a}\Rightarrow \overline{)\begin{array}{c}b=-a{\alpha }^3-a\alpha \end{array}}$$$$\alpha \beta =\alpha {\alpha }^3={\alpha }^4=\frac{c}{a}\Rightarrow \overline{)\begin{array}{c}c=a{\alpha }^4\end{array}}$$$${(b^2-2ac)}^2=ac{(a+c)}^2$$$$\Rightarrow {({(-a{\alpha }^3-a\alpha )}^2-2a\left(a{\alpha }^4\right))}^2=a\left(a{\alpha }^4\right){(a+a{\alpha }^4)}^2$$$${\Rightarrow (a^2{({\alpha }^3+\alpha )}^2-2a^2{\alpha }^4))}^2=a^2\left(a^2{\alpha }^4\right){(1+{\alpha }^4)}^2$$$${\Rightarrow a^4(({\alpha }^6+2{\alpha }^4+{\alpha }^2)-2{\alpha }^4))}^2=\left(a^4{\alpha }^4\right){(1+{\alpha }^4)}^2$$$${\Rightarrow a^4\left(({\alpha }^6+{\alpha }^2)\right))}^2=a^4{\alpha }^4{(1+{\alpha }^4)}^2$$$$\Rightarrow a^4({\alpha }^{12}+2{\alpha }^8+{\alpha }^4=a^4{\alpha }^4(1+2{\alpha }^4+{\alpha }^8)$$$$\Rightarrow a^4({\alpha }^{12}+2{\alpha }^8+{\alpha }^4)=a^4({\alpha }^4+2{\alpha }^8+{\alpha }^{12})$$$$\mathcal{H}enceproved$$

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