Cubic Roots

Let $y=\sqrt[3]{3x-2}$ . Then we have $3y=x^3+2$ and (\3x=y^3+2). Subtracting these two equations, we get $3(y-x)=x^3-y^3=(x-y)(x^2+xy+y^2)$ , which gives $(x-y)(x^2+xy+y^2+3)=0$ . Note that $x^2+xy+y^2+3=\frac{3(x+y)^2}{4}+\frac{(x-y)^2}{4}+3\geq 3>0$ . Thus, $x=y$ and $x=\sqrt[3]{3x-2}$ , which gives $x^3-3x+2=(x-1)^2(x+2)=0$ . Therefore, $x=1$ and $x=-2$ are the two distinct real roots of the given equation.

Consider the substitution $y = \frac {x^3+2}{3}$ . Then, we get $\sqrt[3]{3x-2} = y$ , or that $x= \frac {y^3 + 2}{3}$ . Consider the function $f(a) = \frac {a^3 + 2}{3}$ , which is a monotonically increasing function. We have $f(x) = y$ and $f(y) = x$ , hence $f(f(x))=f(y)=x$ . If $f(x) > x$ , then $ff(x) > f(x) > x$ . Likewise, if $f(x) < x$ , then $ff(x) < f(x) < x$ . Hence, we must have $f(x) = x$ , which means that $\frac {x^3+2} {3} = x$ . This is the cubic $0 = x^3 - 3x+2 = (x-1)^2 (x+2)$ , which has real roots $x=1, -2$ . A quick check shows that both of these roots do indeed satisfy the original equation. Hence, there are 2 distinct real roots to the equation.

First isolate the radical and then cube both sides. A polynomial of degree 9 should result. Then use synthetic division trial and error to find that 1 is a double root and -2 is a single root. The leftover polynomial has no more real factors, so there are 2 distinct real roots.