= Solution
For <equilateral triangular pipe flow>, the <polynomial> vanishes on all three walls, so it satisfies <no-slip boundary conditions>. Direct differentiation gives
$$
\nabla_\perp^2\bigl[(x^2-3y^2)(x-a)\bigr]=(6x-2a)+(-6x+6a)=4a.
$$
Thus
$$
\boxed{A=\frac{G}{4a\mu},\qquad G=\frac{dp}{dz}.}
$$
For $G<0$, $A<0$ and the <velocity> is positive in the triangular interior.
At the wall $x=a$, the outward normal of the <fluid> is $\mathbf e_x$. The axial <traction> exerted by the wall on the <fluid> is $\mu w_x$, so the <fluid> exerts the opposite axial <traction> on the wall. Since $w_x(a,y)=A(a^2-3y^2)$, the <force> per unit axial length is
$$
\frac{\mathbf F}{\text{axial length}}=-\mu A\mathbf k\int_{-a/\sqrt3}^{a/\sqrt3}(a^2-3y^2)\,dy
=-\mu A\frac{4a^3}{3\sqrt3}\mathbf k
=\boxed{-\frac{a^2}{3\sqrt3}\frac{dp}{dz}\,\mathbf k.}
$$
It points with the flow for a negative axial <pressure gradient>. As a check, the triangular <area> is $a^2/\sqrt3$, so the total <pressure> driving <force> per unit length is $-Ga^2/\sqrt3$; each of the three equivalent walls receives one third.
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