Question

Si x+y+2z=0
Calcula A= (x^3+y^3+8z^3)/6xyz

Answer 1

.
$Si x+y+2z=0 \\ Calcula \\ A= \frac{ (x^3+y^3+8z^3)}{6xyz} \\ \\ apicamos \: productos \: notables \: \\ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} ) \\ tambien \: exponentes \: \\ 8 {z}^{3} = {2}^{3} {z}^{3} = {(2z)}^{3} \\ \\ \\ solucion \: \\ \\ en \: A = \frac{(x + y)( {x}^{2} - xy + {y}^{2} )+ (2z)^3 }{6xyz} \\ \\ en \: la \: cndicion \\ \\ Si x+y+2z=0 \\ x + y = - 2z \\ reemplazamos \: en \: A \\ \\ A = \frac{(x + y)( {x}^{2} - xy + {y}^{2} )+ (x + y)^3 }{6xyz} \\ \\ factorizando \: (x + y) \\ \\ A = \frac{(x + y)(( {x}^{2} - xy + {y}^{2} )+ (x + y)^2 )}{6xyz} \\ \\ aplicando \: producto \: notable \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} \\ \\reemplazando en \: A = \frac{(x + y)(( {x}^{2} - xy + {y}^{2} )+ ({x}^{2} + 2xy + {y}^{2}))}{6xyz} \\ \\ operando \: la \: suma \: y \: resta \: de \: terminos \: semejantes \: se \: tendra \: \\ A = \frac{(x + y)(2{x}^{2} + xy +2 {y}^{2} )}{6xyz} \\ \\ aplicamos \: la \: condicion \: despejada \: \\ A = \frac{( - 2z)(2{x}^{2} + xy +2 {y}^{2} )}{6xyz} \\ \\ simplificando \: \\ A = \frac{( - 2z)(2{x}^{2} + xy +2 {y}^{2} )}{2 \times 3xyz} \\ A = \frac{( - 1)(2{x}^{2} + xy +2 {y}^{2} )}{ 3xy} \\ \\ en \: la \: condicion \: elevamos \: al \: cuadrado {()}^{2} \\ \\ {(x + y) }^{2} = {(- 2z ) }^{2} \: \\ {x }^{2} + 2xy + {y}^{2} = 4 {z}^{2} \\ {x}^{2} + {y}^{2} = 4 {z}^{2} - 2xy \: reemplazar \: en \: A \\ A = \frac{( - 1)(2{x}^{2} + xy +2 {y}^{2} )}{ 3xy} \\ A = \frac{( - 1)(2({x}^{2} + {y}^{2} ) + xy)}{ 3xy} \\ A = \frac{( - 1)(2(4 {z}^{2} - 2xy ) + xy)}{ 3xy} \\ \\ A = \frac{( - 1)(8 {z}^{2} - 3xy )}{ 3xy} \\ A = \frac{3xy - 8 {z}^{2} }{3xy} \\ A = \frac{3xy}{3xy} - \frac{8 {z}^{2}}{3xy} \\ A = 1 - \frac{8 {z}^{2}}{3xy}$