Solution: Let the roots of $ f(x) = x^4 - 6x^3 + 11x^2 - 6x + 1 $ be $ r_1, r_2, r_3, r_4 $. We are asked to find $ \sum_{i=1}^4 \frac{1}{r_i} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \frac{1}{r_4} $. Using the identity $ \sum \frac{1}{r_i} = \frac{\sum_{i<j<k} r_i r_j r_k}{r

["Title: Efficiently Compute the Sum of Reciprocals of Roots: A Problem on Polynomial $ f(x) = x^4 - 6x^3 + 11x^2 - 6x + 1 $", "Meta Description: Learn how to compute $ \sum_{i=1}^4 \frac{1}{r_i} $ for the roots $ r_1, r_2, r_3, r_4 $ of the quartic polynomial $ f(x) = x^4 - 6x^3 + 11x^2 - 6x + 1 $ using algebraic identities and Vieta’s formulas.", "---", "### Introduction", "Given the quartic polynomial\n$$\nf(x) = x^4 - 6x^3 + 11x^2 - 6x + 1,\n$$\nlet its roots be $ r_1, r_2, r_3, r_4 $. We aim to compute the sum\n$$\n\sum_{i=1}^4 \frac{1}{r_i} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \frac{1}{r_4}.\n$$", "This expression is symmetric in the roots and can be evaluated efficiently using Vieta’s formulas, avoiding the need to explicitly find the roots.", "---", "### Step 1: Use Vieta’s Formulas", "For a monic polynomial\n$$\nf(x) = x^4 + a_3 x^3 + a_2 x^2 + a_1 x + a_0,\n$$\nVieta’s formulas relate the coefficients to symmetric sums of roots:", "- $ r_1 + r_2 + r_3 + r_4 = -a_3 $\n- $ r_1r_2 + r_1r_3 + \cdots + r_3r_4 = a_2 $\n- $ r_1r_2r_3 + r_1r_2r_4 + \cdots + r_2r_3r_4 = -a_1 $\n- $ r_1r_2r_3r_4 = a_0 $", "Rewriting $ f(x) $ in standard form:\n$$\nf(x) = x^4 - 6x^3 + 11x^2 - 6x + 1,\n$$\nwe identify the coefficients:\n- $ a_3 = -6 $\n- $ a_2 = 11 $\n- $ a_1 = -6 $\n- $ a_0 = 1 $", "From this, Vieta’s formulas give:", "- $ \sum r_i = -(-6) = 6 $\n- $ \sum_{i<j<k} r_i r_j r_k = -(-6) = 6 $\n- $ \prod r_i = 1 $", "---", "### Step 2: Express the Sum of Reciprocals Using Symmetric Sums", "We want:\n$$\n\sum_{i=1}^4 \frac{1}{r_i} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \frac{1}{r_4} = \frac{r_2 r_3 r_4 + r_1 r_3 r_4 + r_1 r_2 r_4 + r_1 r_2 r_3}{r_1 r_2 r_3 r_4}\n$$", "The numerator is $ \sum_{i<j<k} r_i r_j r_k $ — the sum of products of roots taken three at a time — and the denominator is $ r_1 r_2 r_3 r_4 $, the product of all roots.", "From Vieta:", "- Numerator: $ \sum_{i<j<k} r_i r_j r_k = 6 $\n- Denominator: $ r_1 r_2 r_3 r_4 = 1 $", "Thus,\n$$\n\sum_{i=1}^4 \frac{1}{r_i} = \frac{6}{1} = 6\n$$", "---", "### Step 3: Why This Works — The Power of Vieta", "This technique leverages the symmetry and structure encoded in Vieta’s formulas, turning a potential numerical or algebraic computation into a clean, symbolic evaluation. By focusing on symmetric functions of the roots (sums and products), we avoid directly solving for $ r_i $, especially useful for higher-degree polynomials.", "---", "### Final Answer", "$$\n\sum_{i=1}^4 \frac{1}{r_i} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} + \frac{1}{r_4} = 6\n$$", "---", "### Why This Matters", "Understanding how to compute reciprocals of roots using polynomial identities is valuable in algebra, complex analysis, and control theory — especially when analyzing system stability or response to inverse dynamics. Efficiently deriving such expressions saves time and reduces error in complex computations.", "---", "Keywords:\nsum of reciprocals of roots, polynomial roots, Vieta’s formulas, symmetric sums, algebra, polynomial $ f(x) = x^4 - 6x^3 + 11x^2 - 6x + 1 $, $ \sum \frac{1}{r_i} $, symmetric functions of roots, mathematical identity.", "For more insights on polynomial root properties, explore symmetric sums, elementary symmetric polynomials, and Vieta’s connections in algebra curricula and advanced problem solving."]