Integers represented by the polynomial $a^2+b^3+c^6$ Can every sufficiently large integer be written in the form $a^2 + b^3 + c^6 $ for some non-negative integers $a,b$ and $c$?
 A: The number of positive integer triples $(a,b,c)$ such that  $a^2+b^3+c^6\le x$ can be estimated from above by
$$x\int_0^1 \int_0^{(1-c^6)^{1/3}}\int_0^{(1-b^3-c^6)^{1/2}} 1 \, da\, db\, dc\, ,$$ 
that is $\theta x$ with $\theta <1$, so that a portion of integers with positive density is not representable.
A: This is question 3 on page 146 of the second edition of The Hardy-Littlewood Method by R. C. Vaughan, with $b \geq 0.$ 

          


                (Image added by J.O'Rourke)
The answer is No, by a simple volume argument. It is not even necessary to know the exact constant, just that the number of lattice points with $x,y,z \geq 0  $ and $$  x^2 + y^3 + z^6 \leq N$$ is less than $CN,$ with a constant $0 < C < 1.$
I'm in question 5 on the same page. 
Image hosting did not work. Please see: VAUGHAN PDF 
Oh, well. The relevant calculation is the sum of the reciprocals of the exponents, in case the polynomial is the sum of distinct monomials in the different variables. You might think that $$  x^2 + y^2 + z^9$$ ought to represent all large numbers with $z \geq 0.$ There are no local obstructions. But this is not the case. We can think of the exponent in this volume calcultaion as $10/9.$
It is reasonable to ask, how large an exponent reciprocal sum can we get and still fail to represent large integers? The best I have is $4/3,$ and in the simplest form we require coefficients, as in
$$  x^2 + 27 y^2 + 7 z^3.  $$ This is a version of the $ 4 x^2 + 2 x y + 7 y^2 + z^3,  $ which is more natural but looks less diagonal.
