nhan y khac 0 vao pt duoi ta co $28x^2y^2+42xy=7y^3$lay pt tren tru pt ban dau ta co
$-8x^3t^3+28x^2y^2+42xy-27=0=>xy=9/2=>y=\sqrt[3]{756}=>x=\frac3{\sqrt[3]{224}}$
hoac$xy=\frac12=>y=\sqrt[3]{28}=>x=\frac1{2\sqrt[3]{28}}$hoac $xy=-\frac32=>y=\frac{\sqrt[3]{243}}2$