I'm trying to solve the following PDE using Fourier transform
$$\left(1-\triangle\right)^2u=f$$
where $f$ is an integrable function.
I've started with writing it as $$u-2\triangle u+\triangle^2 u=f$$ Using Fourier transform $$\hat u(\textbf{k})\left[1+2\left(k_1^2+k_2^2+k_3^2\right)+\left(?\right)\right]=\hat f(\textbf{k})$$ Where $(?)$ should be replaced by Fourier transform of the biharmonic operator $\triangle^2$.
I found that $$\triangle^2u=u_{xxxx}+u_{xxyy}+u_{xxzz}+u_{yyxx}+u_{yyyy}+u_{yyzz}+u_{zzxx}+u_{zzyy}+u_{zzzz}$$ I know that the transform of $u_{xxxx}$ is $k_1^4\hat u(\textbf{k})$ and similar for $u_{yyyy},u_{zzzz}$, but what should I do with $u_{xxyy}$ for example? Will the transform be $k_1^2k_2^2\hat u(\textbf{k})$?
If so, I get that $$\mathcal{F}\left[\triangle^2 u\right]=\left(k_1^2+k_2^2+k_3^2\right)^2\hat u(\textbf{k})$$ and finally the equation becomes $$\hat u(\textbf{k})\left[1+(k_1^2+k_2^2+k_3^2)\right]^2=\hat f(\textbf{k})$$ In the solution it is written that the transform of $(1-\triangle)^2u$ is $\left(1+\textbf{k}^2\right)^2\hat u(\textbf{k})$, where $\textbf{k}^2=k_1^2+k_2^2+k_3^2$. Could someone please check whether I'm on the right track?
Thanks.