Is it true that $(a^2-ab+b^2)(c^2-cd+d^2)=h^2-hk+k^2$ for some coprime $h$ and $k$? 
Let us consider two numbers of the form $a^2 - ab + b^2$ and $c^2 - cd + d^2$ which are not both divisible by $3$ and such that $(a, b) = 1$ and $(c,d) = 1$. Running some computations it seems that the product $$(a^2 -ab + b^2)(c^2 - cd + d^2) $$ is still of the form $h^2 - hk + k^2$ for some suitable coprime integers $h,k$. Is this true? 

I tried to prove it by writing down explicitly the product and looking for patterns, but I had no luck. Any help would be appreciated!

There is this Identity:
$[(ac+bd)^2-(ab(c^2+d^2)-(abcd)+cd(a^2+b^2))+(bc-ad)^2]=(a^2-ab+b^2)(c^2-cd-d^2)$
Hence for:
$(a^2-ab+b^2)(c^2-cd-d^2)=(h^2-hk+k^2)$
$h=(ac+bd)$
$k=(bc-ad)$
$hk=(ac+bd)(bc-ad)$
Condition (c,d)=(2b,b-2a)
For $(a,b,c,d)=(3,7,14,1)$  we get:
$(49^2-49*95+95^2)=(37)*(183)=6771$
