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Let $a,$ $b,$ $c$ be the roots of $x^3 - x + 4 = 0.$ Compute \[(a^2 - 1)(b^2 - 1)(c^2 - 1).\]

 Aug 19, 2021
 #1
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+1
 Aug 19, 2021
 #2
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0

i tried number one and it was wrong

Penguin  Aug 19, 2021
edited by Penguin  Aug 19, 2021
 #3
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Let \(a,~b,~c\) be the roots of \(x^3 - x + 4 = 0\).
Compute \((a^2 - 1)(b^2 - 1)(c^2 - 1)\).

 

By Vieta:

\(\begin{array}{|rccc|} \hline x^3 &\underbrace{+0*x^2}_{0=-(a+b+c)} &\underbrace{-1*x}_{-1=ab+ac+bc} & \underbrace{+4}_{4=-abc} \\\\ \hline \end{array} \begin{array}{|rcll|} \hline abc &=& -4 \\ ab+ac+bc&=& -1 \\ a+b+c &=& 0 \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline a+b+c &=& 0 \\ (a+b+c)^2 &=& 0^2 \\ a^2+b^2+c^2+2*(ab+ac+bc) &=& 0 \\ a^2+b^2+c^2+2*(-1) &=& 0 \\ \mathbf{a^2+b^2+c^2} &=& \mathbf{2} \\ \hline \end{array} \\ \begin{array}{|rcll|} \hline ab+ac+bc &=& -1 \\ (ab+ac+bc)^2 &=& (-1)^2 \\ a^2b^2+a^2c^2+b^2c^2+2*(a^2bc+ab^2c+abc^2) &=& 1 \\ a^2b^2+a^2c^2+b^2c^2+2*abc*(a+b+c) &=& 1 \\ a^2b^2+a^2c^2+b^2c^2+2*abc*(0) &=& 1 \\ \mathbf{a^2b^2+a^2c^2+b^2c^2} &=& \mathbf{1} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline && \mathbf{(a^2 - 1)(b^2 - 1)(c^2 - 1)} \\\\ &=& a^2b^2c^2-(a^2b^2+a^2c^2+b^2c^2)+a^2+b^2+c^2-1 \\ &=& (-4)^2-(1)+2-1 \\ &=& 16-2+2 \\ &=& \mathbf{16} \\ \hline \end{array}\)

 

\((a^2 - 1)(b^2 - 1)(c^2 - 1) = \mathbf{16}\)

 

laugh

 Aug 19, 2021

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