If I consider the Sobolev space $H^2(\mathbb{R}^3)$ I have the norm 
$$\Vert u\Vert\_{H^2(\mathbb{R}^3)}=\bigg(\sum\_{|\alpha|\leq 2}\Vert D^\alpha u\Vert^2\_{L^2(\mathbb{R}^3)}\bigg)^\frac{1}{2}.$$
Is this norm is equivalent to
$$\Vert u\Vert=\left(\Vert u\Vert\_{L^2}^2+\Vert\Delta u\Vert\_{L^2}^2\right)^{\frac{1}{2}}?$$
In this way if I have to show that a function is in $H^2(\mathbb{R}^3)$ it suffices to study the function and its Laplacian,doesn't it?



Doubt on norm of the Sobolev space $H^2(\mathbb{R}^3)$
You should write:
$$\Vert u\Vert=\left(\Vert u\Vert\_{L^2}^2+\Vert\Delta u\Vert\_{L^2}^2\right)^{\frac{1}{2}}.$$
Then the two norms are equivalent. You can see this by either using interpolation inequalities or by going to the Fourier transform where the estimates are easy:
$$\Vert u\Vert\_{H^2(\mathbb{R}^3)}=\bigg(\sum\_{|\alpha|\leq 2}\Vert |\xi^{\alpha}|\hat u\Vert^2\_{L^2(\mathbb{R}^3)}\bigg)^\frac{1}{2}$$
and 
$$\Vert u\Vert=\left(\Vert \hat u\Vert\_{L^2}^2+\Vert |\xi|^2 \hat u\Vert\_{L^2}^2\right)^{\frac{1}{2}}.$$ 


