Respuesta :

DᴀʀᴋPᴀʀᴀᴅᴏx

[tex] \large \mathfrak{Solution : }[/tex]

Let's factorise :

  • [tex]x( {x}^{2} - 9x + 18) = 0[/tex]

  • [tex]x( {x}^{2} - 6x - 3x + 18) = 0[/tex]

  • [tex]x( x(x - 6) - 3(x - 6)) = 0[/tex]

  • [tex]x(x - 6)(x - 3) = 0[/tex]

now there are three cases :

1. when, x - 6 = 0

  • x = 6

2. when x - 3 = 0

  • x = 3

3. when x = 0

  • x = 0

i hope it helped...

Answer:

x=0,3,6

Step-by-step explanation:

x³-9x²+18x=0

x(x²-9x+18)=0

x[x²-3x-6x+18]=0

x[x(x-3)-6(x-3)]=0

x(x-3)(x-6)=0

x=0,3,6

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