Tìm $x$ biết ${x^3} - 12{x^2} + 48x - 64 = 0$

${x^3} - 3xy + 3{x^2}y + {y^3}$

${x^3} - 6{x^2}y + 12x{y^2} - 8{y^3}$

${x^3} - 6{x^2}y + 12x{y^2} - 4{y^3}$

${x^3} - 3{x^2}y + 12x{y^2} - 8{y^3}$

${\left( {2x + 9} \right)^3}$.

${\left( {2x + 3} \right)^3}$.

${\left( {4x + 3} \right)^3}$.

${\left( {4x + 9} \right)^3}$.

${\left( {2x - y} \right)^3}$.

${\left( {x - 2y} \right)^3}$.

${\left( {4x - y} \right)^3}$.

${\left( {2x + y} \right)^3}$.

${\left( {3x} \right)^3} - {16^3}$.

$9{x^3} - 64$.

$3{x^3} - {4^3}$.

${\left( {3x} \right)^3} - {4^3}$.

${y^3} + {6^3}$.

${\left( {\dfrac{y}{2}} \right)^3} + {6^3}$.

${\left( {\dfrac{y}{2}} \right)^3} + {36^3}$.

${\left( {\dfrac{y}{4}} \right)^3} + {6^3}$.

$x =  - 3$.

$x =  - \dfrac{1}{3}$.

$x = 3$.

$x = \dfrac{1}{3}$.