Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

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]

$$\left.\mathrm{2}\right){x}^{\mathrm{3}} −{qx}−{r}=\mathrm{0} \\ $$$${x}^{\mathrm{3}} +\mathrm{0}×{x}^{\mathrm{2}} +\left(−{q}\right){x}+\left(−{r}\right)=\mathrm{0} \\ $$$$\alpha+\beta+\gamma=\mathrm{0} \\ $$$$\alpha\beta+\beta\gamma+\alpha\gamma=−{q} \\ $$$$\alpha\beta\gamma={r} \\ $$$$\left(\alpha+\beta+\gamma\right)^{\mathrm{2}} =\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} +\mathrm{2}\left(\alpha\beta+\alpha\gamma+\beta\gamma\right) \\ $$$$\mathrm{0}=\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} \right)+\mathrm{2}\left(−{q}\right) \\ $$$${so}\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} =\mathrm{2}{q} \\ $$$${formula}\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{3}{abc}\:{when}\:{a}+{b}+{c}=\mathrm{0} \\ $$$${so} \\ $$$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} =\mathrm{3}\alpha\beta\gamma\:=\mathrm{3}{r}\:\:\left[{since}\:\alpha+\beta+\gamma=\mathrm{0}\right] \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18

c)x^3 =qx+r α^3 =qα+r mjltiply bothside by α^2 α^5 =qα^3 +rα^2

$$\left.{c}\right){x}^{\mathrm{3}} ={qx}+{r} \\ $$$$\alpha^{\mathrm{3}} ={q}\alpha+{r} \\ $$$${mjltiply}\:{bothside}\:{by}\:\alpha^{\mathrm{2}} \\ $$$$\alpha^{\mathrm{5}} ={q}\alpha^{\mathrm{3}} +{r}\alpha^{\mathrm{2}} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18

α^5 =qα^3 +rα^2 β^5 =qβ^3 +rβ^2 γ^5 =qγ^3 +rγ^2 now add them.. 6(α^5 +β^5 +γ^5 ) =6q(α^3 +β^3 +γ^3 )+6r(α^2 +β^2 +γ^2 ) =6q(3r)+6r(2q)=30rq RHS =5(α^3 +β^3 +γ^3 )(α^2 +β^2 +γ^2 ) =5(3r)(2q)=30rq

$$\alpha^{\mathrm{5}} ={q}\alpha^{\mathrm{3}} +{r}\alpha^{\mathrm{2}} \\ $$$$\beta^{\mathrm{5}} ={q}\beta^{\mathrm{3}} +{r}\beta^{\mathrm{2}} \\ $$$$\gamma^{\mathrm{5}} ={q}\gamma^{\mathrm{3}} +{r}\gamma^{\mathrm{2}} \\ $$$${now}\:{add}\:{them}.. \\ $$$$\mathrm{6}\left(\alpha^{\mathrm{5}} +\beta^{\mathrm{5}} +\gamma^{\mathrm{5}} \right) \\ $$$$=\mathrm{6}{q}\left(\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} \right)+\mathrm{6}{r}\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} \right) \\ $$$$=\mathrm{6}{q}\left(\mathrm{3}{r}\right)+\mathrm{6}{r}\left(\mathrm{2}{q}\right)=\mathrm{30}{rq} \\ $$$${RHS}\:=\mathrm{5}\left(\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{3}} \right)\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} +\gamma^{\mathrm{2}} \right) \\ $$$$\:\:\:\:=\mathrm{5}\left(\mathrm{3}{r}\right)\left(\mathrm{2}{q}\right)=\mathrm{30}{rq} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com