Question

Sean $x$ y $y$ números reales que satisfacen $x^4 y^5 + y^4 x^5 = 810$ y $x^3 y^6 + y^3 x^6 = 945.$ Evalúa $2x^3 + (xy)^3 + 2y^3.$

Let $x$ and $y$ be real numbers satisfying $x^4 y^5 + y^4 x^5 = 810$ and $x^3 y^6 + y^3 x^6 = 945.$ Evaluate $2x^3 + (xy)^3 + 2y^3.$

Respuesta

Solución:

Las ecuaciones se factorizan como $x^4y^4(x + y) = 810$ y $x^3y^3(x^3 + y^3) = 945.$ Con $s = x + y$ y $p = xy,$ usando $x^3 + y^3 = s(s^2 - 3p),$ se vuelven $p^4 s = 810$ y $p^3 s\,(s^2 - 3p) = 945.$ Dividiendo, $\begin{aligned} &\frac{s^2 - 3p}{p} = \frac{945}{810} \\ &= \frac{7}{6}, \end{aligned}$ así que $6s^2 = 25p.$

Sustituyendo $p = \frac{6s^2}{25}$ en $p^4 s = 810$ da $\left(\frac{6}{25}\right)^4 s^9 = 810,$ así que $s^9 = 810 \cdot \frac{390625}{1296} = \frac{1953125}{8},$ lo que significa $s^3 = \frac{125}{2}.$ Entonces $ps = \frac{6s^3}{25} = 15$ y $p^3 = \frac{216 s^6}{25^3} = \frac{216 \cdot 15625/4}{15625} = 54.$

Finalmente $\begin{aligned} &2x^3 + (xy)^3 + 2y^3 \\ &= 2(s^3 - 3ps) + p^3 \\ &= 2\left(\frac{125}{2} - 45\right) + 54 \\ &= 35 + 54 = 89. \end{aligned}$

The equations factor as $x^4y^4(x + y) = 810$ and $x^3y^3(x^3 + y^3) = 945.$ With $s = x + y$ and $p = xy,$ using $x^3 + y^3 = s(s^2 - 3p),$ they become $p^4 s = 810$ and $p^3 s\,(s^2 - 3p) = 945.$ Dividing, $\begin{aligned} &\frac{s^2 - 3p}{p} = \frac{945}{810} \\ &= \frac{7}{6}, \end{aligned}$ so $6s^2 = 25p.$

Substituting $p = \frac{6s^2}{25}$ into $p^4 s = 810$ gives $\left(\frac{6}{25}\right)^4 s^9 = 810,$ so $s^9 = 810 \cdot \frac{390625}{1296} = \frac{1953125}{8},$ which means $s^3 = \frac{125}{2}.$ Then $ps = \frac{6s^3}{25} = 15$ and $p^3 = \frac{216 s^6}{25^3} = \frac{216 \cdot 15625/4}{15625} = 54.$

Finally $\begin{aligned} &2x^3 + (xy)^3 + 2y^3 \\ &= 2(s^3 - 3ps) + p^3 \\ &= 2\left(\frac{125}{2} - 45\right) + 54 \\ &= 35 + 54 = 89. \end{aligned}$

2015 AIME II Problema 14

Solución:

El Problema 14 en otros años