Sure, let's solve the given expression step-by-step. The expression provided is:
[tex]\[
\frac{2^{a+3} \cdot 3^{2a-b} \cdot 5^{a+b+3} \cdot 6^{b+1}}{6^{a+1} \cdot 10^{b+3} \cdot 15^a}
\][/tex]
1. Prime Factorization of Composite Numbers:
- \( 6 = 2 \cdot 3 \)
- \( 10 = 2 \cdot 5 \)
- \( 15 = 3 \cdot 5 \)
2. Simplify the Numerator:
- \( 6^{b+1} = (2 \cdot 3)^{b+1} = 2^{b+1} \cdot 3^{b+1} \)
Therefore the numerator becomes:
[tex]\[
2^{a+3} \cdot 3^{2a-b} \cdot 5^{a+b+3} \cdot 2^{b+1} \cdot 3^{b+1}
\][/tex]
Combine the powers of the same base:
[tex]\[
2^{(a+3) + (b+1)} \cdot 3^{(2a-b) + (b+1)} \cdot 5^{a+b+3}
\][/tex]
[tex]\[
2^{a+b+4} \cdot 3^{2a+1} \cdot 5^{a+b+3}
\][/tex]
3. Simplify the Denominator:
- \( 6^{a+1} = (2 \cdot 3)^{a+1} = 2^{a+1} \cdot 3^{a+1} \)
- \( 10^{b+3} = (2 \cdot 5)^{b+3} = 2^{b+3} \cdot 5^{b+3} \)
- \( 15^a = (3 \cdot 5)^a = 3^a \cdot 5^a \)
Therefore the denominator becomes:
[tex]\[
2^{a+1} \cdot 3^{a+1} \cdot 2^{b+3} \cdot 5^{b+3} \cdot 3^a \cdot 5^a
\][/tex]
Combine the powers of the same base:
[tex]\[
2^{(a+1) + (b+3)} \cdot 3^{(a+1) + a} \cdot 5^{(b+3) + a}
\][/tex]
[tex]\[
2^{a+b+4} \cdot 3^{2a+1} \cdot 5^{a+b+3}
\][/tex]
4. Combine Numerator and Denominator:
Now, the expression is:
[tex]\[
\frac{2^{a+b+4} \cdot 3^{2a+1} \cdot 5^{a+b+3}}{2^{a+b+4} \cdot 3^{2a+1} \cdot 5^{a+b+3}}
\][/tex]
5. Simplify the Expression:
Since the numerator and the denominator are identical, the expression simplifies to:
[tex]\[
1
\][/tex]
Thus, the simplified value of the given expression is:
[tex]\[
\boxed{1}
\][/tex]
