1)4x-x³=02)-20x²+5x⁴=03)4x³-3x²=4x-3Помогите если можно то с обеспечением

Решите задачу:

$1)4x - {x}^{3} = 0 \\ x \times (4 - {x}^{2} ) = 0 \\ x = 0 \\ 4 - {x}^{2} = 0 \\ x = 0 \\ x = - 2 \\ x = 2 \\ x1 = - 2 \\ x2 = 0 \\ x3 = 2$
$3)4 {x}^{3} - 3 {x}^{2} = 4x - 3 \\ {x}^{2} \times (4x - 3) = 4x - 3 \\ {x}^{2} \times (4x - 3) - (4x - 3) = 0 \\ (4x - 3) \times ( {x}^{2} - 1) = 0 \\ 4x - 3 = 0 \\ {x}^{2} - 1 = 0\\ x1 = - 1 \\ x2 = 0.75 \\ x3 = 1$

$2) - 20 {x}^{2} + 5 {x}^{4} = 0 \\ - 5 {x}^{2} \times (4 - {x}^{2} ) = 0 \\ {x}^{2} \times (4 - {x}^{2} ) = 0 \\ {x}^{2} = 0 \\ 4 - {x}^{2} = 0 \\ x1 = - 2 \\ x2 = 0 \\ x3 = 2$