Question $\frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}=1$

Studdy Solution

STEP 1

1. $a, b, c$ are non-zero real numbers.
2. The equation is a sum of three fractions set equal to 1.
3. We will need to find a relationship or values for $a, b,$ and $c$ that satisfy the equation.

STEP 2

1. Simplify each term in the equation.
2. Analyze the equation for possible simplifications or substitutions.
3. Solve for $a, b,$ and $c$ .

STEP 3

Simplify each term by rewriting the inverses as fractions:
$\frac{1}{1+a+b^{-1}} = \frac{1}{1+a+\frac{1}{b}}$ $\frac{1}{1+b+c^{-1}} = \frac{1}{1+b+\frac{1}{c}}$ $\frac{1}{1+c+a^{-1}} = \frac{1}{1+c+\frac{1}{a}}$

STEP 4

Consider substituting variables or finding common denominators to simplify the equation further. Here, let's assume a possible symmetry or relationship between $a, b,$ and $c$ .
Assume $a = b = c$ , then each term becomes:
$\frac{1}{1+a+\frac{1}{a}} = \frac{1}{1+a+\frac{1}{a}}$
Since all terms are equal, the equation becomes:
$3 \times \frac{1}{1+a+\frac{1}{a}} = 1$

STEP 5

Solve for $a$ when $a = b = c$ :
$\frac{1}{1+a+\frac{1}{a}} = \frac{1}{3}$
Cross-multiply to solve for $a$ :
$3 = 1 + a + \frac{1}{a}$

STEP 6

Rearrange and solve the equation:
$3 = 1 + a + \frac{1}{a}$ $2 = a + \frac{1}{a}$
Multiply through by $a$ to clear the fraction:
$2a = a^2 + 1$
Rearrange into a standard quadratic form:
$a^2 - 2a + 1 = 0$
Factor the quadratic:
$(a - 1)^2 = 0$
Solve for $a$ :
$a = 1$
Since we assumed $a = b = c$ , we have $a = b = c = 1$ .
The solution is $a = 1, b = 1, c = 1$ .

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