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Question Number 48193 by peter frank last updated on 20/Nov/18

In the equation ax^2 +bx+c=0  one root is square of   orther.without solving  the equation.prove that  c(a−b)^3 =a(c−b)^3

$${In}\:{the}\:{equation}\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0} \\ $$$${one}\:{root}\:{is}\:{square}\:{of}\: \\ $$$${orther}.{without}\:{solving} \\ $$$${the}\:{equation}.{prove}\:{that} \\ $$$${c}\left({a}−{b}\right)^{\mathrm{3}} ={a}\left({c}−{b}\right)^{\mathrm{3}} \\ $$

Answered by MJS last updated on 20/Nov/18

a(x−α)(x−α^2 )=ax^2 −aα(α+1)x+aα^3   aα^3 (a+aα(α+1))^3 =a^4 α^3 (α^2 +α+1)^3   a(aα^3 +aα(α+1))^3 =a^4 α^3 (α^2 +α+1)^3

$${a}\left({x}−\alpha\right)\left({x}−\alpha^{\mathrm{2}} \right)={ax}^{\mathrm{2}} −{a}\alpha\left(\alpha+\mathrm{1}\right){x}+{a}\alpha^{\mathrm{3}} \\ $$$${a}\alpha^{\mathrm{3}} \left({a}+{a}\alpha\left(\alpha+\mathrm{1}\right)\right)^{\mathrm{3}} ={a}^{\mathrm{4}} \alpha^{\mathrm{3}} \left(\alpha^{\mathrm{2}} +\alpha+\mathrm{1}\right)^{\mathrm{3}} \\ $$$${a}\left({a}\alpha^{\mathrm{3}} +{a}\alpha\left(\alpha+\mathrm{1}\right)\right)^{\mathrm{3}} ={a}^{\mathrm{4}} \alpha^{\mathrm{3}} \left(\alpha^{\mathrm{2}} +\alpha+\mathrm{1}\right)^{\mathrm{3}} \\ $$

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