Question

find the value of x in the given equation x^3-12x^2+36x-32=0??

Answer 1

${x}^{3} - 2 {x}^{2} - 10 {x}^{2} + 20x + 16x - 32 = 0$
X^2(X-2)-10X(X-2)+16(X-2)=0

(X-2)(X^2-10X+16)=0

(X-2)(X^2 -8X-2X+16)=0

(X-2)( X-8)(X-2)=0

X=2,8

$x = 2 \: or \: 8$

mark it brainliest