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We can write the following homogenous polynomial of degree $3$ in $3$ variables $$p(a,b,c)=4a^3+18a^2b+28ab^2+15b^3+12a^2c+36abc +28b^2c+12ac^2 +18bc^2+ 4c^3$$ also as sum of cubes of linear forms: $$p(a,b,c)=\left(\sqrt[3]{2}a+\sqrt[3]{\frac{15}{2}-\frac{41}{6\sqrt{3}}}b+\sqrt[3]{2}c\right)^3+\left(\sqrt[3]{2}a+\sqrt[3]{\frac{15}{2}+\frac{41}{6\sqrt{3}}}b+\sqrt[3]{2}c\right)^3$$

I would like to know if there are other sums of cubes of linear forms for this polynomial. If yes which ones? Up to $3$ summed cubes of linear forms are of interest.

An approximate version of the polynomial given here was found by an optimization process that is described in: P. Comon, M. Mourrain: Decomposition of quantics in sums of powers of linear forms, Signal Processing 53, p.93, 1996

https://doi.org/10.1016/0165-1684(96)00079-5

granular_bastard
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1 Answers1

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Proposed polynomial $$p(a,b,c)=4a^3+18a^2b+28ab^2+15b^3+12a^2c+36abc +28b^2c+12ac^2 +18bc^2+4c^3$$ is symmetric by $a, c.$

Let $$s=a+c,\quad p=ac,\quad r=2s+3b,$$ then $$q(s,p,c) = 4s(s^2-3p)+12sp+18b(s^2-2p)+36bp+28b^2s+15b^3$$ $$=4s^3+18s^2b+28sb^2+15b^3=\dfrac12\big((2s+3b)^3+2sb^2+3b^3\big) =\dfrac12(r^3+b^2r)= Q(r,b).$$ There are some different ways to get presentation in cubes.

  • $$Q(r,b)=\dfrac14\left(2r^3+2b^2r\right)=\dfrac14\left(\left(r+\dfrac b{\sqrt3}\right)^3+\left(r-\dfrac b{\sqrt3}\right)^3\right),$$ $$p(a,b,c)=\left(\sqrt[\large3]2a+\dfrac1{\sqrt[\large3]4}\left(3+\dfrac1{\sqrt3}\right)b+\sqrt[\large3]2c\right)^3$$ $$+\left(\sqrt[\large3]2a+\dfrac1{\sqrt[\large3]4}\left(3-\dfrac1{\sqrt3}\right)b+\sqrt[\large3]2c\right)^3.$$
  • $$Q(r,b)=\dfrac16(3r^3+3b^2r) =\dfrac16\left(\left(r+\dfrac2{\sqrt6}\,b\right)^3+2\left(r-\dfrac1{\sqrt6}\,b\right)^3-\dfrac1{\sqrt6}\,b^3\right),$$ $$p(a,b,c)=\left(\sqrt[\large3]{\dfrac43}\,a +\left(\sqrt[\large 3]{\dfrac92}+\dfrac{\sqrt[\large6]6}3\right)b +\sqrt[\large3]{\dfrac43}\,c\right)^3$$ $$+\left(\dfrac2{\sqrt[\large3]3}\,a +\left(\sqrt[\large 3]9 -\dfrac{\sqrt[\large6]{24}}3\right)b +\dfrac2{\sqrt[\large3]3}\,c\right)^3 +\left(-\dfrac b{\sqrt[\large3]6}\right)^3.$$
  • $$Q(r,b)=\dfrac18(4r^3+4r)=\dfrac18\left((r+b)^3+3\left(r-\dfrac b3\right)^3 -\dfrac89 b^3\right),$$ $$p(a,b,c)=(a+2b+c)^3+\left(\sqrt[\large3]3\,a+\dfrac4{\sqrt[\large3]9}b+\sqrt[\large3]3c\right)^3+\left(-\dfrac b{\sqrt9}\right)^3,$$ etc.

Alternative variant over complex numbers (hinted by Will Jagi)

  • $$Q(r,b) = \dfrac12 r(r-ib)(r+ib) = \dfrac12(v+w)(w+u)(u+v),$$ where $u=\dfrac r2,\;v=\dfrac r2+ib,\;w=\dfrac r2-ib.$ $$Q(r,b) = \dfrac16\big((u+v+w)^3 - u^3-v^3-w^3\big) = \dfrac1{48}\big((3r)^3-r^3-(r+2ib)^3-(r-2ib)^3\big)$$ $$=\dfrac{13}{24}r^3-\dfrac1{48}(r+2ib)^3-\dfrac1{48}(r-2ib)^3,$$ $$p(a,b,c) = \left(\sqrt[\large3]{\dfrac{13}{24}}(2a+3b+2c)\right)^3$$ $$+\left(-\dfrac{2a+(3+2i)b+2c}{2\sqrt[\large3]6} \right)^3+\left(-\dfrac{2a+(3-2i)b+2c}{2\sqrt[\large3]6} \right)^3.$$
Yuri Negometyanov
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