Let's break down and simplify the expression step by step.

The given expression is:

$\frac{(5ab^3)^2 \div 0.5}{25(5ab^3)} + 0.7ab^3 - \frac{b^3a}{10}$

Step 1: Simplify the first term inside the fraction.

$(5ab^3)^2 = (5^2)(a^2)(b^6) = 25a^2b^6$

Now, the fraction becomes:

$\frac{25a^2b^6 \div 0.5}{25(5ab^3)}$

Step 2: Divide $25a^2b^6$ by $0.5$.

Dividing by 0.5 is the same as multiplying by 2:

$25a^2b^6 \div 0.5 = 50a^2b^6$

So now the fraction is:

$\frac{50a^2b^6}{25(5ab^3)}$

Step 3: Simplify the denominator.

The denominator is $25 \times 5ab^3 = 125ab^3$.

Now, the fraction becomes:

$\frac{50a^2b^6}{125ab^3}$

Step 4: Simplify the fraction.

Simplify by canceling out common terms:

$\frac{50a^2b^6}{125ab^3} = \frac{50}{125} \times \frac{a^2}{a} \times \frac{b^6}{b^3} = \frac{2}{5} \times a \times b^3$

So, the first term simplifies to:

$\frac{2}{5}ab^3$

Step 5: Combine all the terms.

Now substitute back into the full expression:

$\frac{2}{5}ab^3 + 0.7ab^3 - \frac{b^3a}{10}$

Step 6: Simplify the terms.

All terms have the common factor $ab^3$, so factor them out:

$ab^3 \left(\frac{2}{5} + 0.7 - \frac{1}{10} \right)$

Step 7: Combine the constants.

First, express all terms with a common denominator (denominator of 10):

$\frac{2}{5} = \frac{4}{10}, \quad 0.7 = \frac{7}{10}, \quad \frac{1}{10} = \frac{1}{10}$

Now combine the terms inside the parentheses:

$\frac{4}{10} + \frac{7}{10} - \frac{1}{10} = \frac{10}{10} = 1$

Final Step: The simplified expression.

$ab^3 \times 1 = ab^3$

Thus, the expression simplifies to:

$ab^3$

Would you like further details or have any questions?

Here are 5 related questions to explore:

1. How do you handle more complex algebraic fractions?
2. Can the expression be simplified further using different methods?
3. What are some common mistakes when simplifying expressions like this?
4. How do you deal with exponents when simplifying algebraic terms?
5. How can we apply factoring techniques in complex expressions?

Tip: Always look for common factors when simplifying algebraic expressions; it makes the process more efficient.