u-3-v-3-a-s-3-m-3-b-uv-s-m-2-sm-u-v-2-Find-u-v-s-m-in-terms-of-a-b-

Question Number 75593 by ajfour last updated on 13/Dec/19

u^{3} + v^{3} = a, s^{3} + m^{3} = b,

u v = {(s + m)}^{2}, s m = {(u + v)}^{2} .

F i n d u, v, s, m i n t e r m s o f a, b .

Answered by behi83417@gmail.com last updated on 13/Dec/19

(s + m) [{(s + m)}^{2} - 3 sm] = b \Rightarrow

uv {[uv - 3 {(u + v)}^{2}]}^{2} = b^{2}

(u + v) [{(u + v)}^{2} - 3 uv] = a

\underset{uv = q}{\overset{u + v = p}{\Rightarrow}} {\begin{cases} q {(q - {3 p}^{2})}^{2} = b^{2} \\ p (p^{2} - 3 q) = a \end{cases}

\Rightarrow {\begin{cases} q^{3} - {6 q}^{2} p^{2} + {9 qp}^{4} = b^{2} \\ p^{3} - 3 pq = a \Rightarrow q = \frac{p^{3} - a}{3 p} \end{cases}

\Rightarrow \frac{{(p^{3} - a)}^{3}}{{27 p}^{3}} - {6 p}^{2} \frac{{(p^{3} - a)}^{2}}{{9 p}^{2}} + {9 p}^{4} \frac{p^{3} - a}{3 p} = b^{2}

\Rightarrow {(p^{3} - a)}^{3} - {18 p}^{3} {(p^{3} - a)}^{2} + {81 p}^{6} (p^{3} - a) = {27 p}^{3} b^{2}

\overset{p^{3} = t}{\Rightarrow} {(t - a)}^{3} - 18 t {(t - a)}^{2} + {81 t}^{2} (t - a) = {27 tb}^{2}

\Rightarrow (t^{3} - {3 at}^{2} + {3 a}^{2} t - a^{3}) - 18 t (t^{2} - 2 at + a^{2}) + {81 t}^{2} (t - a) = {27 tb}^{2}

\Rightarrow {64 t}^{3} - {48 at}^{2} - ({15 a}^{2} + {27 b}^{2}) t - a^{3} = 0

\Rightarrow t^{3} - \frac{3}{4} {at}^{2} - \frac{3}{64} ({5 a}^{2} + {9 b}^{2}) t - \frac{a^{3}}{64} = 0

\overset{T = t + \frac{a}{4}}{\Rightarrow} T^{3} - \frac{3}{64} ({17 a}^{2} + {9 b}^{2}) T - \frac{3}{256} ({4 a}^{3} + {5 a}^{2} + {9 b}^{2}) = 0

△ = - 3 \times \frac{- 3}{256} ({4 a}^{3} + {5 a}^{2} + {9 b}^{2}) > 0

k = \frac{- 27 \times \frac{- 3}{256} ({4 a}^{3} + {5 a}^{2} + {9 b}^{2})}{2 \times \frac{9}{256} ({4 a}^{3} + {5 a}^{2} + {9 b}^{2}) \sqrt{\frac{9}{256} ({4 a}^{3} + {5 a}^{2} + {9 b}^{2})}} < 1

T_{1} = \frac{\sqrt{{4 a}^{3} + {5 a}^{2} + {9 b}^{2}}}{8} \times \cos (\frac{1}{3} \cos^{- 1} \frac{24}{\sqrt{{4 a}^{3} + {5 a}^{2} + {9 b}^{2}}})

T_{2} = \frac{\sqrt{{4 a}^{3} + {5 a}^{2} + {9 b}^{2}}}{8} \times \cos (\frac{1}{3} \cos^{- 1} \frac{24}{\sqrt{{4 a}^{3} + {5 a}^{2} + {9 b}^{2}}} - \frac{2 π}{3})

T_{3} = \frac{\sqrt{{4 a}^{3} + {5 a}^{2} + {9 b}^{2}}}{8} \times \cos (\frac{1}{3} \cos^{- 1} \frac{24}{\sqrt{{4 a}^{3} + {5 a}^{2} + {9 b}^{2}}} + \frac{2 π}{3})

p_{i}^{3} = t_{i} \overset{i = 1, 2, 3}{=} T_{i} - \frac{a}{4} = \frac{\sqrt{{4 a}^{3} + {5 a}^{2} + {9 b}^{2}}}{8} \cos (\frac{1}{3} \cos^{- 1} \frac{24}{\sqrt{{4 a}^{3} + {5 a}^{2} + {9 b}^{2}}}) - \frac{a}{4}

\Rightarrow p_{1} = \sqrt[3]{\frac{\sqrt{{4 a}^{3} + {5 a}^{2} + {9 b}^{2}}}{8} \cos (\frac{1}{3} \cos^{- 1} \frac{24}{\sqrt{{4 a}^{3} + {5 a}^{2} + {9 b}^{2}}}) - \frac{a}{4}}

q_{1} = \frac{p^{3} - a}{3 p} = \frac{\frac{\sqrt{{4 a}^{3} + {5 a}^{2} + {9 b}^{2}}}{8} \cos (\frac{1}{3} \cos^{- 1} \frac{24}{\sqrt{{4 a}^{3} + {5 a}^{2} + {9 b}^{2}}}) - \frac{5 a}{4}}{3 \sqrt[3]{\frac{\sqrt{{4 a}^{3} + {5 a}^{2} + {9 b}^{2}}}{8} \cos (\frac{1}{3} \cos^{- 1} \frac{24}{\sqrt{{4 a}^{3} + {5 a}^{2} + {9 b}^{2}}}) - \frac{a}{4}}}

u = - \frac{p}{2} + \frac{\sqrt{p^{2} - 4 q}}{2}, v = - \frac{p}{2} - \frac{\sqrt{p^{2} - 4 q}}{2}

s = - \frac{uv}{2} + \frac{\sqrt{uv - 4 {(u + v)}^{2}}}{2}

m = - \frac{uv}{2} - \frac{\sqrt{uv - 4 {(u + v)}^{2}}}{2}