Question Number 49354 by peter frank last updated on 06/Dec/18
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18
![2)x^3 −qx−r=0 x^3 +0×x^2 +(−q)x+(−r)=0 α+β+γ=0 αβ+βγ+αγ=−q αβγ=r (α+β+γ)^2 =α^2 +β^2 +γ^2 +2(αβ+αγ+βγ) 0=(α^2 +β^2 +γ^2 )+2(−q) so α^2 +β^2 +γ^2 =2q formula a^3 +b^3 +c^3 =3abc when a+b+c=0 so α^3 +β^3 +γ^3 =3αβγ =3r [since α+β+γ=0]](https://www.tinkutara.com/question/Q49419.png)
$$\left.\mathrm{2}\right){x}^{\mathrm{3}} −{qx}−{r}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} +\mathrm{0}×{x}^{\mathrm{2}} +\left(−{q}\right){x}+\left(−{r}\right)=\mathrm{0} \\ $$$$\alpha+\beta+\gamma=\mathrm{0} \\ $$$$\alpha\beta+\beta\gamma+\alpha\gamma=−{q} \\ $$$$\alpha\beta\gamma={r} \\ $$$$\left(\alpha+\beta+\gamma\right)^{\mathrm{2}} =\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} +\mathrm{2}\left(\alpha\beta+\alpha\gamma+\beta\gamma\right) \\ $$$$\mathrm{0}=\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} \right)+\mathrm{2}\left(−{q}\right) \\ $$$${so}\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} =\mathrm{2}{q} \\ $$$${formula}\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{3}{abc}\:{when}\:{a}+{b}+{c}=\mathrm{0} \\ $$$${so} \\ $$$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} =\mathrm{3}\alpha\beta\gamma\:=\mathrm{3}{r}\:\:\left[{since}\:\alpha+\beta+\gamma=\mathrm{0}\right] \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18
$$\left.{c}\right){x}^{\mathrm{3}} ={qx}+{r} \\ $$$$\alpha^{\mathrm{3}} ={q}\alpha+{r} \\ $$$${mjltiply}\:{bothside}\:{by}\:\alpha^{\mathrm{2}} \\ $$$$\alpha^{\mathrm{5}} ={q}\alpha^{\mathrm{3}} +{r}\alpha^{\mathrm{2}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18
$$\alpha^{\mathrm{5}} ={q}\alpha^{\mathrm{3}} +{r}\alpha^{\mathrm{2}} \\ $$$$\beta^{\mathrm{5}} ={q}\beta^{\mathrm{3}} +{r}\beta^{\mathrm{2}} \\ $$$$\gamma^{\mathrm{5}} ={q}\gamma^{\mathrm{3}} +{r}\gamma^{\mathrm{2}} \\ $$$${now}\:{add}\:{them}.. \\ $$$$\mathrm{6}\left(\alpha^{\mathrm{5}} +\beta^{\mathrm{5}} +\gamma^{\mathrm{5}} \right) \\ $$$$=\mathrm{6}{q}\left(\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} \right)+\mathrm{6}{r}\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} \right) \\ $$$$=\mathrm{6}{q}\left(\mathrm{3}{r}\right)+\mathrm{6}{r}\left(\mathrm{2}{q}\right)=\mathrm{30}{rq} \\ $$$${RHS}\:=\mathrm{5}\left(\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} \right)\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} \right) \\ $$$$\:\:\:\:=\mathrm{5}\left(\mathrm{3}{r}\right)\left(\mathrm{2}{q}\right)=\mathrm{30}{rq} \\ $$$$ \\ $$