Can $x^3+x^2+1=0$ be solved using high school methods?

Question

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?

Answer 1

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.

Answer 2

Here is the cubic solving process, and in this case you can get to the lone real root without using complex numbers.

$$x^3+x^2+1=0$$

Eliminate the quadratic term with a substitution: $x=y-\frac13$.

$$\begin{align} 0=x^3+x^2+1 &=\left(y-\frac13\right)^3+\left(y-\frac13\right)^2+1\\ &=y^3-y^2+\frac13y-\frac1{27}+y^2-\frac23y+\frac19+1\\ &=y^3-\frac13y+\frac{29}{27}\\ \end{align}$$

Now imagine the potential existence of numbers $u$ and $v$ such that $y=u+v$: $$\begin{align} 0 &=(u+v)^3-\frac13(u+v)+\frac{29}{27}\\ &=u^3+3uv(u+v)+v^3-\frac13(u+v)+\frac{29}{27}\\ &=u^3+\left(3uv-\frac13\right)(u+v)+v^3+\frac{29}{27}\\ \end{align}$$

Now imagine that these two numbers $u$ and $v$ just happen to make $3uv-\frac13=0$. Soon enough we will see that such $u$ and $v$ do indeed exist. In this case we have

$$\begin{align} 0 &=u^3+v^3+\frac{29}{27}\\ \end{align}$$

And since we are assuming $3uv-\frac13=0$, that means that $v=\frac{1}{9u}$.

$$\begin{align} 0 &=u^3+\left(\frac{1}{9u}\right)^3+\frac{29}{27}\\ \implies 0 &=u^6+\frac{272}{27}u^3+\frac{1}{9^3}\\ \implies u^3&=\frac{-\frac{29}{27}\pm\sqrt{\left(\frac{29}{27}\right)^2-4\left(\frac{1}{9^3}\right)}}{2} \end{align}$$

Note that either the solution for $u^3$ with $+$ or the one with $-$ is a real number, since the number in the radical is positive. Since all we wanted is a $u$ and $v$ that meet the conditions we laid out, we can go ahead and take $$u=\sqrt[3]{\frac{-\frac{29}{27}+\sqrt{\left(\frac{29}{27}\right)^2-4\left(\frac{1}{9^3}\right)}}{2}}$$ and let $v=\frac{1}{9u}$, and this pair of $u$ and $v$ meets our condition that $3uv-\frac13=0$.

With these laid out, we can back substitute to $y=u+v$. And then to $x=y-\frac13$. In the end,

$$x=\sqrt[3]{\frac{-\frac{29}{27}+\sqrt{\left(\frac{29}{27}\right)^2-4\left(\frac{1}{9^3}\right)}}{2}}+\frac{1}{9\sqrt[3]{\frac{-\frac{29}{27}+\sqrt{\left(\frac{29}{27}\right)^2-4\left(\frac{1}{9^3}\right)}}{2}}}-\frac13$$

(Incidentally, it can be shown that $v^3$ is actually the other root from the quadratic-solving step, so

$$x=\sqrt[3]{\frac{-\frac{29}{27}+\sqrt{\left(\frac{29}{27}\right)^2-4\left(\frac{1}{9^3}\right)}}{2}}+\sqrt[3]{\frac{-\frac{29}{27}-\sqrt{\left(\frac{29}{27}\right)^2-4\left(\frac{1}{9^3}\right)}}{2}}-\frac13$$

is another way to express the root.)

Answer 3

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 $$x=u+b={1\over 3}(-1+m+n)={1\over 3}\left(-1+\sqrt[3]{\frac{-29+ 3\sqrt{93}}{2}}+\sqrt[3]{\frac{-29- 3\sqrt{93}}{2}}\right)$$

Answer 4

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.