Let $E$ be given by $y^2 = x^3 + Ax + B$ over a field $K$ and let $d \in K^\times$. The twist of $E$ by $d$ is the elliptic curve $E^{(d)}$ given by $y^2 = x^3 + Ad^2x + Bd^3$.
Show that $j(E^{(d)}) = j(E)$
I know $j(E) = 1728 \frac{4A^3}{4A^3 + 27B^2}$ and that a twist is when two different elliptic curves of the same field have the same j-invariant.
The book is super vague, but it mentions some kind of change of base where $A_1 = \mu^4A$ and $B_1 = \mu^6B$, but I don't understand it or how to apply it.
I'm completely lost, any help would be amazing.
You will get the result by applying the formula for $j(E)$ for the new coefficients $A' := Ad^2$ and $B' := Bd^3$.