Question

If a,b,c are roots of the equation x3 + px2 + qx + r = 0, form the equation whose roots are
​a-1/bc, b-1/ca, c- 1/ab​

Answer 1

Dear student
$$\begin{gathered} Ifa,bandcaretherootsofacubicpolynomialf\left(x\right),then \\ f\left(x\right)=k\left\{x^3-\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x-abc\right\} \\ wherekisanynon-zerorealnumber. \\ Here, \\ a+b+c=\frac{a-1}{bc}+\frac{b-1}{ca}+\frac{c-1}{ab} \\ \Rightarrow a+b+c=\frac{a\left(a-1\right)+b\left(b-1\right)+c\left(c-1\right)}{abc} \\ \Rightarrow \mathit{a}\mathbf{+}\mathit{b}\mathbf{+}\mathit{c}\mathbf{=}\frac{\left({\mathbf{a}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}^{\mathbf{2}}\right)\mathbf{-}\left(\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{+}\mathbf{c}\right)}{\mathbf{a}\mathbf{b}\mathbf{c}} \\ and,ab+bc+ca=\left[\left(\frac{a-1}{bc}\right)\left(\frac{b-1}{ca}\right)+\left(\frac{b-1}{ca}\right)\left(\frac{c-1}{ab}\right)+\left(\frac{c-1}{ab}\right)\left(\frac{a-1}{bc}\right)\right] \\ \Rightarrow ab+bc+ca=\frac{ab\left(ab-a-b+1\right)+bc\left(bc-b-c+1\right)+ca\left(ca-c-a+1\right)}{a^2{b}^2{c}^2} \\ \Rightarrow \mathit{a}\mathit{b}\mathbf{+}\mathit{b}\mathit{c}\mathbf{+}\mathit{c}\mathit{a}\mathbf{=}\frac{\left({\mathbf{a}}^{\mathbf{2}}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}{\mathbf{c}}^{\mathbf{2}}\mathbf{+}{\mathbf{c}}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{2}}\right)\mathbf{-}{\mathbf{a}}^{\mathbf{2}}\left(\mathbf{b}\mathbf{+}\mathbf{c}\right)\mathbf{-}{\mathbf{b}}^{\mathbf{2}}\left(\mathbf{c}\mathbf{+}\mathbf{a}\right)\mathbf{-}{\mathbf{c}}^{\mathbf{2}}\left(\mathbf{a}\mathbf{+}\mathbf{b}\right)\mathbf{+}\left(\mathbf{a}\mathbf{b}\mathbf{+}\mathbf{b}\mathbf{c}\mathbf{+}\mathbf{c}\mathbf{a}\right)}{{\mathbf{a}}^{\mathbf{2}}{\mathbf{b}}^{\mathbf{2}}{\mathbf{c}}^{\mathbf{2}}} \\ and,\mathit{a}\mathit{b}\mathit{c}\mathbf{=}\left(\frac{\mathbf{a}\mathbf{-}\mathbf{1}}{\mathbf{b}\mathbf{c}}\right)\left(\frac{\mathbf{b}\mathbf{-}\mathbf{1}}{\mathbf{c}\mathbf{a}}\right)\left(\frac{\mathbf{c}\mathbf{-}\mathbf{1}}{\mathbf{a}\mathbf{b}}\right) \\ so,thepolynomialis \\ f\left(x\right)=k\left\{x^3-\left(\frac{\left({\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2\right)-\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)}{\mathrm{abc}}\right)x^2+\left(\frac{\left({\mathrm{a}}^2{\mathrm{b}}^2+{\mathrm{b}}^2{\mathrm{c}}^2+{\mathrm{c}}^2{\mathrm{a}}^2\right)-{\mathrm{a}}^2\left(\mathrm{b}+\mathrm{c}\right)-{\mathrm{b}}^2\left(\mathrm{c}+\mathrm{a}\right)-{\mathrm{c}}^2\left(\mathrm{a}+\mathrm{b}\right)+\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)}{{\mathrm{a}}^2{\mathrm{b}}^2{\mathrm{c}}^2}\right)x-\left(\frac{\mathrm{a}-1}{\mathrm{bc}}\right)\left(\frac{\mathrm{b}-1}{\mathrm{ca}}\right)\left(\frac{\mathrm{c}-1}{\mathrm{ab}}\right)\right\} \end{gathered}$$