# x1,x2,x3 are the roots of the equation f(x)=x^3-3x^2+4x-2=0 y1=6/f'(x1), y2=6/f"(x2), y3=6/f"(x3) are the roots of the eq. y^3+ay^2+by+c=0Find the value of the parameter "m", knowing that m=a+b+c...

x1,x2,x3 are the roots of the equation

f(x)=x^3-3x^2+4x-2=0

y1=6/f'(x1), y2=6/f"(x2), y3=6/f"(x3) are the roots of the eq.

y^3+ay^2+by+c=0

Find the value of the parameter "m", knowing that m=a+b+c and f:R->R.

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### f(x)=x^3-3x^2+4x-2=(x-1)(x^2 - 2x + 2) = 0

The roots of this equation (x1, x2, x3) are: x=1 and x = 1 +- i

f'(x) = 3x^2 - 6x + 4 --> f'(1) = 1 ; f'(1+-i) = -2

f''(x) = 6x -6 --> f"(1) = 0; f"(1+-i) = +-6i

Note that x1 = 1, since 1/f"(1) blows up. Thus, {x2,x3} = 1 +- i

Thus the roots of y^3 + ay^2 + by + c = zero are:

y1 = 6

y2 = 6/6i = -i

y3 = -6/6i = i

We have three unknowns, a b and c, and three equations:

6^3 + a6^2 + 6b + c = 0 (1)

i^3 + ai^2 + bi + c = 0 (2)

(-i)^3 + a(-i)^2 - bi + c = 0 (3)

Summing (2) and (3) yields: a + c = 0. Thus a + b + c = b = m.

The difference between (2) and (3) yields: b = 1

**Therefore, m = 1**

x^3-3x^2+4x-2. Since for x=1, f(x)=0.Threfore,x-1 is a factor of f(x).

Therefore, f(x)/(x-1) = x^2-2x+2.

Therefore, the roots of f(x) are got from the factors:x-1=0 and the quadratic x^2-2x+2= Therfore, x1=1 and

x2 =1+i or x3 = 1 - i, where i=sqrt(-1).

f'(x)= (x^3+3x^2+2)' =3x^2-6x+4.

f''(x)=(3x^2-6x+4)'=6(x-1)

f'(x1)=3*1^2-4*1+2=1

f''(x2)=6(1+i-1) = 6i

f''(x3) = 6(1-i-1)= -6i

y1 =6/f'(x) =**6,** y2= 6/f''(x2)= 6/[6*i)= 1/i= **-i**

y3=6/f''(x3)=6/(-6i)=**i**

y^3+ay^2+by+c=0.

By the relation of coefficients and the roots,

-a=y1+y2+y3= 6+i-i= 6 or **a=-6**

b=y1y2+y2y3+y3y1=y1(y2+y3)+y2y3=6(0)+(-i)(i) =**1**

-c=6+(-i)+(i)= . So, **c=-6**

Therefore, m=a+b+c= -6+1-6 =** 1**

y^3+ay^2+by+c is ** **y^3**-6**y^2+**1**y **- 6**