Mathematical Proof: Solving for the Product of Ratios of Roots of Polynomial Equations

In this article, we will delve into a detailed mathematical proof involving the roots of polynomial equations. Specifically, we will solve for the product of ratios of the roots of the polynomial equation x^3 - x^2 - 1 0. This problem is an excellent example of how to use algebraic manipulation and coefficient comparison to find a compact and elegant solution.

Introduction

The problem arises from a polynomial equation with roots a, b, and c. Our goal is to find the value of the expression (frac{a}{bc} cdot frac{b}{ca} cdot frac{c}{ab}). To solve this, we will follow a step-by-step approach that involves expressing the polynomial equation in a factored form and comparing coefficients.

Step-by-Step Solution

Given the polynomial equation

x^3 - x^2 - 1 0

We can express it as

(x - a)(x - b)(x - c) x^3 - cx^2 - bx^2 - ax^2 - abx - bcx - acx - abc x^3 - x^2 - 1

By expanding the left-hand side, we obtain

x^3 - (a b c)x^2 (ab bc ca)x - abc x^3 - x^2 - 1

Now, we can compare the coefficients on both sides of the equation to derive useful relationships between the roots.

Step 1: Coefficient Comparison

Comparing the coefficients of x^2, we get

a b c 1

Comparing the coefficients of x, we get

ab bc ca 0

Finally, comparing the constant term, we get

abc 1

Using these relationships, we can proceed to solve the expression (frac{a}{bc} cdot frac{b}{ca} cdot frac{c}{ab}).

Step 2: Simplifying the Expression

Let’s write the expression as a fraction

(frac{a}{bc} cdot frac{b}{ca} cdot frac{c}{ab} frac{a^2b^2c^2}{(abc)^2})

We know from the coefficient comparison that

(abc)^2 a^2b^2c^2

Thus, we have

(frac{a^2b^2c^2}{(abc)^2} frac{a^2b^2c^2 - 2abbc (-2abc)(ab bc ca)}{(abc)^2})

Substituting abc 1 and ab bc ca 0, we get

(frac{a^2b^2c^2 - 2abbc (2ab 2bc 2ca)}{(abc)^2})

frac{1^2 - 2 times 0}{1^2})

frac{1}{1})

Therefore, the value of the expression is 1.

Conclusion

This problem demonstrates the power of algebraic manipulation and coefficient comparison in solving complex polynomial equations. By following a systematic approach, we can derive the value of the given expression efficiently and accurately.