due $$ 81 a^{3} b^{5} c^{5}+27 b^{3} c $$

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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:
   $$
   27b^3c \left(3a^3b^2c^4 + 1\right)
   $$
The expression factors to $$27b^3c(3a^3b^2c^4 + 1)$$.

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

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