Zero of Polynomial

Question

If $(a-b)$, $a$, and $(a+b)$ are the zeroes of the polynomial $x^3-3x^2+x+1$ then what are the values of $a$ and $b$? I have taken $f(x)=x^3-3x^2+x+1$ and equated $f(a-b)$, $f(a)$ and $f(a+b)$ to zero. But I could not reduce the equations.

Answer 1

HINT:

Clearly, $$a-b+a+a+b=\frac31$$

and $$(a-b)a(a+b)=-\frac11$$

Answer 2

Hint: The roots for $(x-x_1)(x-x_2)(x-x_3)$ are $x_1,x_2,x_3$.

Answer 3

So you got the equations

• $a^3 - 3a^2 b + 3ab^2 - b^3 - 3a^2 + 6ab - 6b^2 + a - b + 1 = 0$
• $a^3 - 3a^2 + a + 1 = 0$
• $a^3 + 3a^2 b + 3a b^2 + b^3 - 3a^2 - 6ab - 6b^2 + a + b + 1 = 0$

The challenge is how to systematically simplify these equations. It is too much to try and solve for one variable to plug into the other equations: instead, you want to combine the equations in ways that eliminate the more complicated terms.

I've written the monomials in order of their total degree, and arbitrarily broke ties by putting the highest powers in $a$ first. An approach is to try and eliminate the terms as far to the left when possible when we can.

For example, we can eliminate $a^3$ from the first and third equation by subtracting off the second:

• $-3a^2 b + 3ab^2 - b^3 + 6ab - 6b^2 - b = 0$
• $a^3 - 3a^2 + a + 1 = 0$
• $3a^2 b + 3a b^2 + b^3 - 6ab - 6b^2 + b = 0$

and so forth. Another thing we can do is to factor equations when we can:

• $b(-3a^2 + 3ab - b^2 + 6a - 6b - 1) = 0$
• $a^3 - 3a^2 + a + 1 = 0$
• $b(3a^2 + 3a b + b^2 - 6a - 6b + 1) = 0$

at which point we can make a significant simplification by splitting into two parts: in one of the two parts we add the equation $b=0$, and in the other part we insist on $b \neq 0$, and can divide $b$ out of the equations.

By repeating these steps, you can fairly quickly come upon something you can solve in the ordinary way.