Let $b_1$, $b_2$, $b_3$, $c_1$, $c_2$, and $c_3$ be real numbers such that for every real number $x$, we have
\[
  x^6 - x^5 + x^4 - x^3 + x^2 - x + 1 = (x^2 + b_1 x + c_1)(x^2 + b_2 x + c_2)(x^2 + b_3 x + c_3). 
\]
Compute $b_1 c_1 + b_2 c_2 + b_3 c_3$. 
1. We start with the given polynomial:
   \[
   P(x) = x^6 - x^5 + x^4 - x^3 + x^2 - x + 1
   \]
   We need to factor this polynomial into a product of three quadratic polynomials with real coefficients:
   \[
   P(x) = (x^2 + b_1 x + c_1)(x^2 + b_2 x + c_2)(x^2 + b_3 x + c_3)
   \]

2. The roots of \(P(x)\) are the 7th roots of unity, excluding \(1\). The 7th roots of unity are given by:
   \[
   e^{2k\pi i / 7} \quad \text{for} \quad k = 0, 1, 2, 3, 4, 5, 6
   \]
   Excluding \(1\) (which corresponds to \(k = 0\)), the remaining roots are:
   \[
   e^{2\pi i / 7}, e^{4\pi i / 7}, e^{6\pi i / 7}, e^{8\pi i / 7}, e^{10\pi i / 7}, e^{12\pi i / 7}
   \]

3. These roots can be paired into conjugate pairs to form quadratic factors with real coefficients. The pairs are:
   \[
   (e^{2\pi i / 7}, e^{-2\pi i / 7}), (e^{4\pi i / 7}, e^{-4\pi i / 7}), (e^{6\pi i / 7}, e^{-6\pi i / 7})
   \]

4. Each pair of roots \((e^{2k\pi i / 7}, e^{-2k\pi i / 7})\) forms a quadratic factor:
   \[
   (x - e^{2k\pi i / 7})(x - e^{-2k\pi i / 7}) = x^2 - (e^{2k\pi i / 7} + e^{-2k\pi i / 7})x + 1
   \]
   Using Euler's formula, \(e^{2k\pi i / 7} + e^{-2k\pi i / 7} = 2\cos(2k\pi / 7)\), we get:
   \[
   x^2 - 2\cos(2k\pi / 7)x + 1
   \]

5. Therefore, the quadratic factors are:
   \[
   x^2 - 2\cos\left(\frac{2\pi}{7}\right)x + 1, \quad x^2 - 2\cos\left(\frac{4\pi}{7}\right)x + 1, \quad x^2 - 2\cos\left(\frac{6\pi}{7}\right)x + 1
   \]

6. From these factors, we identify:
   \[
   b_1 = -2\cos\left(\frac{2\pi}{7}\right), \quad b_2 = -2\cos\left(\frac{4\pi}{7}\right), \quad b_3 = -2\cos\left(\frac{6\pi}{7}\right)
   \]
   and
   \[
   c_1 = 1, \quad c_2 = 1, \quad c_3 = 1
   \]

7. We need to compute \(b_1 c_1 + b_2 c_2 + b_3 c_3\):
   \[
   b_1 c_1 + b_2 c_2 + b_3 c_3 = -2\cos\left(\frac{2\pi}{7}\right) \cdot 1 + -2\cos\left(\frac{4\pi}{7}\right) \cdot 1 + -2\cos\left(\frac{6\pi}{7}\right) \cdot 1
   \]
   \[
   = -2\left(\cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right)\right)
   \]

8. Using the identity for the sum of cosines of angles in arithmetic progression:
   \[
   \cos\left(\frac{2\pi}{7}\right) + \cos\left(\frac{4\pi}{7}\right) + \cos\left(\frac{6\pi}{7}\right) = -\frac{1}{2}
   \]

9. Substituting this result, we get:
   \[
   b_1 c_1 + b_2 c_2 + b_3 c_3 = -2 \left(-\frac{1}{2}\right) = 1
   \]

The final answer is \(\boxed{1}\)