How to solve $x^6- 4x^3+4=0$?

Question

I'm working through an exercise with the question:

Use substitution to solve for $x$ in the following equation $x^6 - 4x^3 + 4 = 0$

The solution given is $\sqrt[3]2$ however there is no working in between shown and I'm at a loss for how to arrive at this.

I started trying to factor our the polynomials $x^6$ and $4x^3$ but that didn't really get me anywhere:

$x^3(x^3 - 4) = -4$ is as far as I got.

How can I arrive at $\sqrt[3]2$ with each step shown in between?

Answer 1

Hint: put $y = x^3$, and you get a quadratic equation.

Answer 2

Use that $x^6-4x^3+4=(x^3-2)^2$

Answer 3

A simple substitution gets the job done. Let $t = x^3$.

$$x^3(x^3-4) = -4 \iff t(t-4) = -4 \iff t = 2$$

$$t = 2 \iff x^3 = 2 \iff x = \sqrt[3]{2}$$

As a more general tip, whenever you have “quadratic-like” equations, make a substitution to reach a normal quadratic.

$$ax^{2n}+bx^n+c = 0$$

Here, letting $t = x^n$ reduces the problem to solving for $t$ in $at^2+bt+c = 0$.

Answer 4

Let $ y = x^{3} $. Thus $$x^6 - 4x^3 + 4 $$$$ = (x^{3})^{2} - 4(x^{3}) + 4$$$$ = y^{2} - 4y + 4$$ which is a quadratic equation that can be factored using the perfect squares method.