What is the inverse of: f(vector) = vector*squareMagnitude

Question

I believe the square magnitude of a 3 component vector is: (xx + yy + z*z). If you multiply a vector by its square magnitude, is there a function that you can perform to obtain the original vector?

Lets take original vector3 A = (Ax, Ay, Az). The final vector3 B = A*(AxAx + AyAy + Az*Az).

Now, is there a function of B that will return A?

Thank you!

I originally found that the dot product of a Vector with itself equals the vectors square magnitude. However, I also read that the inverse of dot product is impossible because of many solutions. Still, with the two qualifiers (1: the dot product is between a vector and itself, 2: the final vector's components are linearly related to the original vector's corresponding components) there may be enough to restrict to one possible solution.

Answer 1

Get squared magnitude of B. Note that result SMB is equal to magnitude of A in the sixth power (or cube of SMA=squareMagnitude(A)).

B = bx +       by +       bx = 
    ax * SMA + ay * SMA + az * SMA

SMB = B.dot.B = ax^2 * SMA^2 + ay^2 * SMA^2 + az^2 * SMA^2 = 
                SMA^2*(ax^2 + ay^2 + az^2) = 
                SMA^2 * SMA = 
                SMA^3

So you can get squareMagnitude(A) as cube root of SMB and extract components of A

 SMA = SMB^(1/3) = pow(SBB, 1./3)    
 ax = bx / SMA
 ay = by / SMA
 az = bz / SMA

P.S. It is interesting - what operations do provide B value?

Concerning the second question - not very readable in comments, so I've added this:

A/(1+SMA)=B   square it
SMA/(1+2*SMA+SMA^2)=SMB
SMA = SMB+2*SMA*SMB+SMA^2*SMB
SMA^2*SMB+SMA*(2*SMB-1)+SMB = 0 
D = 4*SMB^2-4*SMB+1-4*SMB^2= 1-4*SMB  
   D should be positive, so B magnitude should be small 
SMA = (1-2*SMB + Sqrt(1-4*SMB)) /(2*SMB)
    the second solution with minus is definitely negative, so omit it
 and finally with known SMA
A = B * (1 + SMA)

Answer 2

B = (|A|^2) A so |B| = |A|^3, so A = B/(|A|^2) = B/(|B|^(2/3)).