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?
 A: 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 $$x=u+b={1\over 3}(-1+m+n)={1\over 3}\left(-1+\sqrt[3]{\frac{-29+ 3\sqrt{93}}{2}}+\sqrt[3]{\frac{-29- 3\sqrt{93}}{2}}\right)$$
A: No there is not - unless of course they have learned about complex numbers as well as the formula for cubic roots.
Depends on the high-school, they may have studied Newton's method, and may be expected to find the single real root numerically.
A: A cubic equation of the form $x^3 + a x^2 + b x + c = 0$ is reduced to an equation that can be solved with the quadratic formula by the substitution $$x = \frac{1}{3}\left(\frac{a^2-3b}{y} + y - a\right).$$
For example in this specific case we have $a=1$ and $b=0$, suggesting the substitution $$x = \frac{1}{3}\left(\frac{1}{y} + y - 1\right)$$ which leads to the equation $$y^6+29 y^3+1 = 0.$$ This can be solved for $y^3$ by the quadratic formula and $x$ is obtained from possible values for $y$ after taking a cube root.
