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1-3-i-2-n-x-i-i-2-n-1-i-j-2-n-j-3-i-1-j-1-x-i-i-j-1-n-x-j-i-1-j-1-j-i-i-j-1-n-j-i-n-3-i-1-




Question Number 195227 by York12 last updated on 28/Jul/23
  α_1 ^3 [((Π_(i=2) ^n (x−α_i ))/(Π_(i=2) ^n (α_1 −α_i )))]+Σ_(j=2) ^n (α_j ^3 [((Π_(i=1) ^(j−1) (x−α_i )Π_(i=j+1) ^n (x−α_j ))/(Π_(i=1) ^(j−1) (α_j −α_i )Π_(i=j+1) ^n (α_j −α_i )))]+α_n ^3 [((Π_(i=1) ^(n−1) (x−α_i ))/(Π_(i=1) ^(n−1) (α_n −α_i )))]−x^3 =0  solve for x .            [ where n≥5 ]
$$ \\ $$$$\alpha_{\mathrm{1}} ^{\mathrm{3}} \left[\frac{\underset{{i}=\mathrm{2}} {\overset{{n}} {\prod}}\left({x}−\alpha_{{i}} \right)}{\underset{{i}=\mathrm{2}} {\overset{{n}} {\prod}}\left(\alpha_{\mathrm{1}} −\alpha_{{i}} \right)}\right]+\underset{{j}=\mathrm{2}} {\overset{{n}} {\sum}}\left(\alpha_{{j}} ^{\mathrm{3}} \left[\frac{\underset{{i}=\mathrm{1}} {\overset{{j}−\mathrm{1}} {\prod}}\left({x}−\alpha_{{i}} \right)\underset{{i}={j}+\mathrm{1}} {\overset{{n}} {\prod}}\left({x}−\alpha_{{j}} \right)}{\underset{{i}=\mathrm{1}} {\overset{{j}−\mathrm{1}} {\prod}}\left(\alpha_{{j}} −\alpha_{{i}} \right)\underset{{i}={j}+\mathrm{1}} {\overset{{n}} {\prod}}\left(\alpha_{{j}} −\alpha_{{i}} \right)}\right]+\alpha_{{n}} ^{\mathrm{3}} \left[\frac{\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\left({x}−\alpha_{{i}} \right)}{\underset{{i}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\prod}}\left(\alpha_{{n}} −\alpha_{{i}} \right)}\right]−{x}^{\mathrm{3}} =\mathrm{0}\right. \\ $$$${solve}\:{for}\:{x}\:.\:\:\:\:\:\:\:\:\:\:\:\:\left[\:{where}\:{n}\geqslant\mathrm{5}\:\right] \\ $$

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