Show directly from the definition of continuity that $f(x)=x^3 +x$ is continuous

Question

My attempt

Let $\epsilon>0$. Setting $\delta = ???, |x-x_{0}|\leqslant \delta\implies$ $|f(x)-f(x_{0})|=|x^3+x-x_{0}^3-x| \leqslant|x^3-x_{0}^3|+|x-x_{0}| = |(x-x_{0})(x+x_{0})^2|+|x-x_{0}|$

From here where do I go, do I let $|x-x_{0}| \leqslant 1$ such that $-1\leqslant x-x_{0} \leqslant 1$ then solve for $x\leqslant 1+x_{0}$ and sub that in such that, $|(x-x_{0})(x+x_{0})^2|+|x-x_{0}| \leqslant |(1+x_{0}-x_{0})(1+x_{0}+x_{0})^2|+|x-x_{0}|=|1+2x_{0}|^2+|x-x_{0}|$

But then I get stuck from here. Any tips would be appreciated!

EDIT Let $\epsilon>0$. Setting $\delta =min(1,\frac{\epsilon}{|1+4x_{0}+2x_{0}^2|+1}), |x-x_{0}|\leqslant \delta\implies$ $|f(x)-f(x_{0})|=|x^3+x-x_{0}^3-x| \leqslant|x^3-x_{0}^3|+|x-x_{0}| = |(x-x_{0})(x^2+xx_{0}+x_{0}^2)|+|x-x_{0}| \leqslant |x-x_{0}||x^2+xx_{0}+x_{0}^2|+|x-x_{0}| \leqslant \delta(|x^2+xx_{0}+x_{0}^2|+1)\leqslant \delta(|(1+x_{0})^2+(1+x_{0})x_{0}+x_{0}^2|+1)=\delta(|1+4x_{0}+2x_{0}^2|+1) < \epsilon$

Is this correct?

Answer 1

Hint:

$$x^3-x_0^3=(x-x_0)(x^2+xx_0+x_0^2)$$

To be used at the very beginning.