To express $\frac{9x^2 + 3}{(x-1)(x^2 + x + 10)}$ as a partial fraction, we proceed by decomposing it as follows:

$\frac{9x^2 + 3}{(x-1)(x^2 + x + 10)} = \frac{A}{x-1} + \frac{Bx + C}{x^2 + x + 10}$

Step 1: Clear the Denominator

Multiply both sides by the common denominator $(x-1)(x^2 + x + 10)$:

$9x^2 + 3 = A(x^2 + x + 10) + (Bx + C)(x - 1)$

Step 2: Expand the Right Side

Expand each term on the right side:

1. Expand $A(x^2 + x + 10)$: $A(x^2 + x + 10) = Ax^2 + Ax + 10A$

2. Expand $(Bx + C)(x - 1)$: $(Bx + C)(x - 1) = Bx^2 - Bx + Cx - C = Bx^2 + (C - B)x - C$

Thus, the right side becomes:

$Ax^2 + Ax + 10A + Bx^2 + (C - B)x - C$

Step 3: Combine Like Terms

Combine terms with $x^2$, $x$, and the constant terms:

$(A + B)x^2 + (A + C - B)x + (10A - C)$

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