# What are x1^2 + x2^2 + x3^2 and x1^3 + x2^3 + x3^3 if f = 2x^3 - 3x^2 + x ?

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We notice that the order of the given polynomial f is 3, so x1, x2, x3 are the roots of the polynomial.

We'll write Viete's relations for a polynomial of 3rd order:

ax^3 + bx^2 + cx + d = 0

x1 + x2 + x3 = -b/a

x1*x2 + x1*x3 + x2*x3 = c/a

x1*x2*x3 = -d/a

We'll identify a,b,c,d:

a = 2

b = -3

c = 1

d = 0

x1 + x2 + x3 = 3/2 (1)

x1*x2 + x1*x3 + x2*x3 = 1/2 (2)

x1^2 + x2^2 + x3^2 = (x1 + x2 + x3)^2 - 2(x1*x2 + x1*x3 + x2*x3) (3)

We'll substitute (1) and (2) in (3):

x1^2 + x2^2 + x3^2 = 9/4 - 2/2 = (9-4)/4 = 5/4

**x1^2 + x2^2 + x3^2 = 5/4**

We also know that if x1,x2,x3 are the roots of f, then if we'll substitute them into f, they will cancel f.

f(x1) = 0

2x1^3 - 3x1^2 + x1 = 0

f(x2) = 0

2x2^3 - 3x2^2 + x2 = 0

f(x3) = 0

2x3^3 - 3x3^2 + x3

f(x1) + f(x2) + f(x3) = 0

2(x1^3+x2^3+x3^3) - 3(x1^2+x2^2+x3^2) + x1 + x2 + x3 = 0

2(x1^3+x2^3+x3^3) = 3(x1^2+x2^2+x3^2) - (x1 + x2 + x3)

2(x1^3+x2^3+x3^3) = 15/4 - 3/2

2(x1^3+x2^3+x3^3) = (15-6)/4

2(x1^3+x2^3+x3^3) = 9/4

We'll divide by 2:

**(x1^3+x2^3+x3^3) = 9/8**

To find x1^2 + x2^2 + x3^2 and x1^3 + x2^3 + x3^3 if f = 2x^3 - 3x^2 + x.

f(x) = 2x^3-3x^2+x.

We presume x1,x2 and x3 are the roots of 2x^3-3x^2+x = 0.

2x^3-3x^2+x = 0

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

x (2x-1)(x- 1) = 0.

x = 0, or 2x-1 = 0, or x-1 = 0

Therefore x1 = 0, x2 = 1/2 and x3 = 1.

Therefore x1^2+x2^2+x3^2 = 0^2 +(1/2)^2+1^2 = 5/4.

x1^3+x2^3+x3^3 = 0^3+(1/2)^3+1^3 = 9/8.