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x2+bx+c is a factor of x3-p. Show that p=b3 (Hint: find that c=b2 first)

 Mar 29, 2021
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Notice that \(x^3-p=x^3-(\sqrt[3]{p})^3=(x-\sqrt[3]{p})(x^2+x\sqrt[3]{p}+(\sqrt[3]{p})^2)\), using the rule of difference of cubes. Since we know that \(x^2+x\sqrt[3]{p}+(\sqrt[3]{p})^2\)is a factor of \(x^3-p\), we now know that \(\sqrt[3]{p}=b\). Cube both sides of the equation to get \(\boxed{p=b^3}\)

 

I genuinely do not understand why the hint would be helpful to solve this problem in any way, but you can show that's true by noticing that \(c=(\sqrt[3]{p})^2=b^2\) 

 Mar 29, 2021

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