due \( 81 a^{3} b^{5} c^{5}+27 b^{3} c \)

1. Identify the greatest common factor (GCF) of the two terms: - The numerical coefficients are \(81\) and \(27\). Their GCF is \(27\). - For the variable \(b\): the powers are \(b^5\) and \(b^3\). The smallest power is \(b^3\). - For the variable \(c\): the powers are \(c^5\) and \(c\). The smallest power is \(c\). - The variable \(a\) appears only in the first term, so it is not common. 2. Factor out the GCF \(27 b^3 c\) from the expression: \[ 81a^3b^5c^5 + 27b^3c = 27b^3c \left(\frac{81a^3b^5c^5}{27b^3c} + \frac{27b^3c}{27b^3c}\right) \] 3. Simplify each term inside the parentheses: - For the first term: \[ \frac{81a^3b^5c^5}{27b^3c} = \frac{81}{27} \cdot a^3 \cdot b^{5-3} \cdot c^{5-1} = 3a^3b^2c^4 \] - For the second term: \[ \frac{27b^3c}{27b^3c} = 1 \] 4. Write the final factored expression: \[ 27b^3c \left(3a^3b^2c^4 + 1\right) \]