Let the function $f(x)=\sqrt[3]{-x}+\sqrt[3]{x^2}+x$ for any real x negative and $f(c)=0$ with $c \in ]-\infty,-1]$

Question

Let the function $$f(x)=\sqrt[3]{-x}+\sqrt[3]{x^2}+x$$ for any real $x$ negative and $f(c)=0$ with $c\in ]-\infty,-1]$

Prove that : $c$ verifies the equation $c^3+4c^2-c=0$

Answer 1

We have $$ f(c)=\sqrt[3]{-c}+\sqrt[3]{c^2}+c = 0\\ \sqrt[3]{-c}+\sqrt[3]{c^2}= -c\\ (\sqrt[3]{-c}+\sqrt[3]{c^2})^3= -c^3\\ -c + 3c\sqrt[3]c -3c\sqrt[3]{c^2} + c^2 = -c^3\\ -c + c^2 -3c(\sqrt[3]{-c} + \sqrt[3]{c^2})= -c^3 $$ Since we assumed that $\sqrt[3]{-c} + \sqrt[3]{c^2} = -c$ (line 2 above), we may just substitute it, and we get $$ -c + c^2 - 3c(-c) = -c^3\\ c^3 + 4c^2 - c = 0 $$

Answer 2

Observe that, substituting $\;t:=\sqrt[3]x\;$ , we have

$$-\sqrt[3]x+\left(\sqrt[3]x\right)^2+x=t^2-t+t^3=t(t^2+t-1)$$

Thus, any root of the above expression in $\;t\;$ different from zero fulfills:

$$t^2+t-1=0\implies t_{1,2}=\frac{-1\pm\sqrt5}{2}\stackrel{\text{Choose the negative}}\implies t=-\frac12\left(1+\sqrt5\right)$$

and now:

$$x=t^3=-2-\sqrt5\;,\;\;\text{so for any such root}\;\;c\;:$$

$$c^3+4x^2-x=(-2-\sqrt5)^3+4(-2-\sqrt5)^2-(-2-\sqrt5)=$$

$$=(-2-\sqrt5)\left[\,4+4\sqrt5+5-8-4\sqrt5-1\,\right]=0$$

Answer 3

By following identity :

$ (x+y+z)(x^2+y^2+z^2-xy-yz-zx)=$$x^3+y^3+z^3 - 3xyz$

set : $x=\sqrt[3]{-c}$ ,$y =\sqrt[3]{c^2}$ , $z=c$

so $x+y+z =0$ ( Because $f(c)=0$)

so $x^3+y^3+z^3 - 3xyz=0$$\Rightarrow$$c^3 + 4c^2 - c = 0$