We are given that $a$, $b$, and $c$ are the roots of the cubic equation:

$x^3 - 6x^2 + 2x - 1 = 0$

This implies that by Vieta's formulas:

• $a + b + c = 6$
• $ab + bc + ca = 2$
• $abc = 1$

We are tasked with finding the polynomial whose roots are:

$\frac{1}{a} + \frac{1}{b}, \quad \frac{1}{b} + \frac{1}{c}, \quad \frac{1}{c} + \frac{1}{a}$

Step 1: Express the new roots

The new roots can be written as:

$\frac{1}{a} + \frac{1}{b} = \frac{a+b}{ab}, \quad \frac{1}{b} + \frac{1}{c} = \frac{b+c}{bc}, \quad \frac{1}{c} + \frac{1}{a} = \frac{c+a}{ca}$

Using Vieta’s formulas:

• $a + b + c = 6$
• $ab + bc + ca = 2$
• $abc = 1$

We can now express the new roots as: $\frac{a+b}{ab} = \frac{6 - c}{1/c} = 6 - c$ $\frac{b+c}{bc} = 6 - a, \quad \frac{c+a}{ca} = 6 - b$

So the new roots are $6 - a$, $6 - b$, and $6 - c$.

Step 2: Finding the new polynomial

Since the roots are $6 - a$, $6 - b$, and $6 - c$, the polynomial with these roots can be obtained by making a substitution $y = 6 - x$ in the original polynomial $P(x)$, where $P(x) = x^3 - 6x^2 + 2x - 1$.

Thus, the new polynomial is $P(6 - y)$. To find this, substitute $x = 6 - y$ into the original polynomial:

$P(6 - y) = (6 - y)^3 - 6(6 - y)^2 + 2(6 - y) - 1$

Now expand and simplify the expression:

1. Expand $(6 - y)^3 = 216 - 108y + 18y^2 - y^3$
2. Expand $(6 - y)^2 = 36 - 12y + y^2$, so $-6(6 - y)^2 = -216 + 72y - 6y^2$
3. Expand $2(6 - y) = 12 - 2y$

Now substitute everything into the equation:

$P(6 - y) = (216 - 108y + 18y^2 - y^3) + (-216 + 72y - 6y^2) + (12 - 2y) - 1$

Simplifying:

$P(6 - y) = -y^3 + (18y^2 - 6y^2) + (-108y + 72y - 2y) + (216 - 216 + 12 - 1)$

$P(6 - y) = -y^3 + 12y^2 - 38y + 11$

Thus, the polynomial whose roots are $\frac{1}{a} + \frac{1}{b}, \frac{1}{b} + \frac{1}{c}, \frac{1}{c} + \frac{1}{a}$ is:

$y^3 - 12y^2 + 38y - 11 = 0$

Would you like any further clarification on the steps involved?

Here are five related questions to consider:

1. What are Vieta's formulas and how do they relate to polynomial roots?
2. How can you determine the sum and product of roots for any cubic polynomial?
3. What is the process of polynomial substitution and when is it useful?
4. How does changing variables in a polynomial affect its roots?
5. How do symmetric expressions of the roots simplify computations?

Tip: When working with roots of polynomials, substitution techniques can simplify finding new polynomials with transformed roots.