(3/8)(3h−p)^2 +3ph=(3h^2 −1) and (((3h−p)^3 )/(16))+p(3h^2 −1)=h^3 −h−c Find p and h in terms of 0<c<(2/(3(√3)))∙


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Question Number 198414 by ajfour last updated on 19/Oct/23

$\frac{3}{8} {(3 h - p)}^{2} + 3 p h = (3 h^{2} - 1)$ $a n d$ $\frac{{(3 h - p)}^{3}}{16} + p (3 h^{2} - 1) = h^{3} - h - c$ $F i n d p a n d h i n t e r m s o f 0 < c < \frac{2}{3 \sqrt{3}} \cdot$
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