Any idea?

Notice that if we let $x=3+\sqrt[3]{3}$ then $(x-3)^3=3\implies x^3+27x=9x^2+30$. Define a rectangle as good when it can be tiled by rectangles that are similar to $1\times x$. Obviously, if $a\times b$ is good then $ac\times bc$ is also good for every positive real numbers $c$. Also, if $m\times n$ is good then $m\times jn$ is good for every positive integer $j$. Claim: If $a\times b$ is good then $ak\times bl$ is also good for every positive rational numbers $k,l$. Proof. Let $k=\frac{r}{s}$ and $l=\frac{p}{q}$ with $p,q,r,s$ as positive integers. Denote $``\to"$ as an operation to indicate that the good configuration at the tail implies that the configuration at the head is also good. $$a\times b \to a\times b\color{blue}p\color{black}\to \frac{a}{\color{blue}q}\color{black}\times \frac{bp}{\color{blue}q}\color{black} \to \left(\frac{a}{q}\cdot \color{blue}qr\color{black}\right)\times \frac{bp}{q} \to ar\times \frac{bp}{q}\to\frac{ar}{\color{blue}s}\color{black}\times \frac{bp}{q\color{blue}s\color{black}}\to \frac{ar}{s}\times \left(\frac{bp}{qs}\cdot\color{blue}s\color{black} \right) \to ak\times bl$$ Note that $$x\times 1\to x\times \color{blue}9\color{black}\to \left(x\cdot\color{blue}\left(x^2+\frac{10}{3}\right)\color{black}\right)\times \left(9 \cdot\color{blue}\left(x^2+\frac{10}{3}\right)\color{black}\right)\to\left(x^3+\frac{10x}{3}\right)\times (9x^2+30x)$$$$1\times \color{blue}x\color{black}\to (1\cdot\color{blue}x\color{black})\times (x\cdot \color{blue}x\color{black})\to x\times x^2\to\color{blue}\frac{71}{3}\color{black}x\times \color{blue}9\color{black}x^2\quad\text{ and }\quad x\times 1\to\color{blue}\frac{71}{3}\color{black}x\times\color{blue}30\color{black}.$$Thus $\frac{71}{3}x\times (9x^2+30)$ is good and so $(x^3+27x)\times (9x^2+30)$ is good. Since $x^3+27x=9x^2+30x$, we have $$(x^3+27x)\times (9x^2+30)\to \frac{x^3+27x}{\color{blue}x^3+27x}\color{black}\times\frac{9x^2+30}{\color{blue}x^3+27x}\color{black}\to \boxed{1\times 1}$$

There is a general result by Laczkovich and Szekeres (1994) about tiling a square with similar rectangles which states that a square can be decomposed into rectangles similar to a rectangle of size $1 \times k$ if and only if $k$ is algebraic and each of its conjugates has positive real part. The conjugates of $(3 + \sqrt[3]{3})$ are the other two roots of its minimal polynomial, $(x-3)^3=3$, (which are $\sqrt[3]{3}\omega+3$ and $\sqrt[3]{3}\omega ^2 + 3$, where $\omega$ is the cube root of unity), whose real parts are positive.