Question

les doy corona si me ayudan en este ejercicio plisss
x³+(−b2−bc) x¹+(b2c+bc2)=0

Answer 1

Explicación paso a paso:

x³+(−b2−bc) x¹+(b2c+bc2)= 0

x^3 + (-2b-bc) x^1 + (2bc +bc .2)= 0

x^3+(-2b-bc)x + (2bc+2bc) = 0

x^3+(-2b-bc)x + b = 0

x^3+x(-2b-bc) +b =0

x(-2b -bc) + b = -x^3

x(-2b -bc) = -x^3 - b

x . -b(2+c)= -x^3-b

-b(2+c)= (-x^3-b)/ x

-b(2+c)= -x^2 - (b/ x)

2+c= - ((-x^2 - b/x)/b)

c= ((-x^2 - b/x)/b) -2