<s>[INST] Can $x^3+x^2+1=0$ be solved using high school methods? I encountered the following problem in a high-school math text, which I wasn't able to solve it: $x^3 + x^2 + 1 = 0$ Am I missing something here, or is indeed a more advanced method necessary to solve this particular cubic?
 [/INST] Let $x=au+b$. Then $$x^3+x^2+1= a^3u^3+3a^2bu^2+3ab^2u+b^3+a^2u^2+2abu+b^2+1$$ $$=a^3u^3+(3a^2b+a^2)u^2+(3ab^2+2ab)u+(b^3+b^2+1)$$We want $3b+1=0\implies b=-1/3$ and then $$b^3+b^2+1=\frac{-1+3+27}{27}=\frac{29}{27}$$ and $$3ab^2+2ab=a(1/3-2/3)=-\frac{a}{3}$$ So clearly denominators we get $$(3au)^3-3(3au)+29=0$$ Now let $m+n=3au$ so that $$m^3+n^3+3mn(m+n)-3(m+n)+29=0$$ or $$m^3+n^3+3(m+n)(mn-1)+29=0$$ and require that $mn=1$ so that we get $m^3+n^3=-29$ and $m^3n^3=1^3=1$. We have two numbers whose sum and product are known, so they solve the quadratic $y^2+29y+1=0$ which has roots (with help from a calculator) $$\frac{-29\pm 3\sqrt{93}}{2}$$ $m$ and $n$ are the cube roots of this. We never needed to specify $a$ so take it as $1$. We finally get $$x=u+b={1\over 3}(-1+m+n)={1\over 3}\left(-1+\sqrt[3]{\frac{-29+ 3\sqrt{93}}{2}}+\sqrt[3]{\frac{-29- 3\sqrt{93}}{2}}\right)$$
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