alpha,beta,gamma are roots of x3-px2+qx-r=0 find value of [beta+ gamma- alpha)*(gamma+alpha-beta)*(alpha+ beta-gamma)
Answers
\underline{\textbf{Given:}}
Given:
\mathsf{\alpha,\beta,\gamma\;are\;roots\;of\;x^3-p\,x^2+q\,x-r=0}α,β,γarerootsofx
3
−px
2
+qx−r=0
\underline{\textbf{To find:}}
To find:
\textsf{The value of}The value of
\mathsf{(\beta+\gamma-\alpha)\;(\gamma+\alpha-\beta)\;(\alpha+\beta-\gamma)}(β+γ−α)(γ+α−β)(α+β−γ)
\underline{\textbf{Solution:}}
Solution:
\mathsf{Since\;\alpha,\beta,\gamma\;are\;roots\;of\;x^3-p\,x^2+q\,x-r=0,\;we\;have}Sinceα,β,γarerootsofx
3
−px
2
+qx−r=0,wehave
\mathsf{\alpha+\beta+\gamma=p}α+β+γ=p
\mathsf{\alpha\beta+\beta\gamma+\gamma\alpha=q}αβ+βγ+γα=q
\mathsf{\alpha\,\beta\,\gamma=r}αβγ=r
\mathsf{Now,}Now,
\mathsf{(\beta+\gamma-\alpha)\,(\gamma+\alpha-\beta)\,(\alpha+\beta-\gamma)}(β+γ−α)(γ+α−β)(α+β−γ)
\mathsf{=(\alpha+\beta+\gamma-2\,\alpha)\,(\gamma+\alpha+\beta-2\,\beta)\,(\alpha+\beta+\gamma-2\,\gamma)}=(α+β+γ−2α)(γ+α+β−2β)(α+β+γ−2γ)
\mathsf{=(p-2\,\alpha)\,(p-2\,\beta)\,(p-2\,\gamma)}=(p−2α)(p−2β)(p−2γ)
\mathsf{=p^3-2(\alpha+\beta+\gamma)\,p^2+4(\alpha\beta+\beta\gamma+\gamma\alpha)\,p-8\,\alpha\beta\gamma}=p
3
−2(α+β+γ)p
2
+4(αβ+βγ+γα)p−8αβγ
\mathsf{=p^3-2(p)p^2+4(q)p-8r}=p
3
−2(p)p
2
+4(q)p−8r
\mathsf{=p^3-2p^3+4pq-8r}=p
3
−2p
3
+4pq−8r
\mathsf{=-p^3+4pq-8r}=−p
3
+4pq−8r
\mathsf{=-p^3+4pq-8r}=−p
3
+4pq−8r
\implies\boxed{\boldmath{(\beta+\gamma-\alpha)\;(\gamma+\alpha-\beta)\;(\alpha+\beta-\gamma)=-p^3+4pq-8r}}⟹
\boldmath(β+γ−α)(γ+α−β)(α+β−γ)=−p
3
+4pq−8r