Question

hola podrian ayudarme con productos notables

Answer 1

recordando :

a³+b³= (a+b)(a²-ab+b²)


piden :

$E= \sqrt{ x^{9}+ y^{9}+3 x^{4}.y^{4} }$

degradando :

x³+y³ = (x+y)(x²-xy+y²)
( xy )³= (x³+y³)³
 
$x^{3}y ^{3}= x^{9 } +y^{9} +3 x^{3}y ^{3} ( x^{3} +y ^{3}) \\ \\ x^{3} .y ^{3} = x^{9} +y ^{9} +3 x^{3} .y ^{3} . x^{3} .y^{3} \\ \\ x^{3} .y ^{3}(1-3 x^{3} .y ^{3})=x^{9 } +y^{9}$

remplazando en  E

$E= \sqrt{ x^{9}+ y^{9}+3 x^{4}.y^{4} } \\ \\ E= \sqrt{ x^{3}.y ^{3}(1-3 x^{3}.y ^{3})+3 x^{4} .y ^{4} } \\ \\ E= \sqrt{ x^{3}.y ^{3}[(1-3x^{3}.y ^{3})+3xy)] } \\ \\ E= \sqrt{ x^{3}.y ^{3}(1-3x^{3}.y ^{3}+3xy )}$
$E= \sqrt{ x^{3}.y ^{3}(1-3x^{3}.y ^{3}+3xy )} \\ \\ xy\ \textgreater \ 0 ahora \\ \\ (1-3xy( x^{2} y^{2} -1\ \textless \ 1 \\ \\ E= \sqrt{ x^{3}.y ^{3}.1 } \\ \\ E= \sqrt{ x^{2} . y^{2}.xy } \\ \\ E=xy \sqrt{xy}$

saludos Isabela