$\tan\alpha, \tan\beta, \tan\gamma$ are the roots of the equation $au^{3}+(20-x)u+y=0$ for fixed $\mathrm{x}$ and $\mathrm{y}$, and $\tan\alpha+\tan\beta=h$, then $ah^{3}+(20-x)h=$

Let the three roots of the cubic au cubed + (20 - x)u + y = 0 be r, s, and t. By Vieta's formulas, the sum of the roots r + s + t equals 0. We are given that r + s = h, which means the third root t must equal -h. Since t = -h is a root of the equation, substituting u = -h into the polynomial must satisfy it. This gives a(-h) cubed + (20 - x)(-h) + y = 0, which simplifies to -ah cubed - (20 - x)h + y = 0. Rearranging this equation yields ah cubed + (20 - x)h = y.

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