x^{2}+bx+c is a factor of x^{3}-p. Show that p=b^{3} (Hint: find that c=b^{2} first)

Guest Mar 29, 2021

#1**0 **

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 \(c=(\sqrt[3]{p})^2=b^2\)

textot Mar 29, 2021