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\alpha^3 + p\alpha^2 + q\alpha + r = 0 \\[br](1-\alpha)^3 + p(1 - \alpha)^2 + q(1-\alpha) + r = 0 \\[br]\dfrac{1}{\alpha^3} + \dfrac{p}{\alpha^2} + \dfrac{q}{\alpha} + r = 0
1 - 3\alpha + 3\alpha^2 - \alpha^3 + p - 2\alpha p + p\alpha^2 + p - p\alpha + 1 = 0 \\[br]\implies \alpha^3 - 1 + 3\alpha - 3\alpha^2 - p + 2\alpha p - p \alpha^2 - p + p\alpha - 1 = 0 \\[br]\implies \alpha^3 + (-3-p)\alpha^2 + (3+3p)\alpha - (2p + 2) = 0
u_{k-1} - u_'{k-1} = 0
u_{k} - u_'{k} = 0
x^2 + bk + c = 0 \\[br]\dfrac{1}{k^2} + \dfrac{b}{k} + c = 0 \implies 1 + bk + ck^2 = 0 \implies k^2 + \dfrac{b}{c}k + \dfrac{1}{c} = 0
\alpha^3 + p\alpha^2 + q\alpha + r = 0 \\[br](1-\alpha)^3 + p(1 - \alpha)^2 + q(1-\alpha) + r = 0 \\[br]\dfrac{1}{\alpha^3} + \dfrac{p}{\alpha^2} + \dfrac{q}{\alpha} + r = 0
1 - 3\alpha + 3\alpha^2 - \alpha^3 + p - 2\alpha p + p\alpha^2 + p - p\alpha + 1 = 0 \\[br]\implies \alpha^3 - 1 + 3\alpha - 3\alpha^2 - p + 2\alpha p - p \alpha^2 - p + p\alpha - 1 = 0 \\[br]\implies \alpha^3 + (-3-p)\alpha^2 + (3+3p)\alpha - (2p + 2) = 0
\displastyle\lim_{x\to\infty}h(x)=0
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