To solve the problem, we need to find the values of \( a \), \( b \), and \( c \) that satisfy the given conditions. Let's break it down step by step.
### Step 1: Understand the properties of G.P.
Given that \( a \), \( b \), and \( c \) are in increasing geometric progression (G.P.), we can express them as:
- \( a = x \)
- \( b = xr \)
- \( c = xr^2 \)
where \( r > 1 \) is the common ratio.
### Step 2: Use the logarithmic equation
We are given:
\[
\log_6 a + \log_6 b + \log_6 c = 6
\]
Using the property of logarithms that states \( \log_b(m) + \log_b(n) = \log_b(mn) \), we can combine the logarithms:
\[
\log_6 (abc) = 6
\]
This implies:
\[
abc = 6^6 = 46656
\]
### Step 3: Express \( abc \) in terms of \( x \) and \( r \)
Substituting \( a \), \( b \), and \( c \) in terms of \( x \) and \( r \):
\[
abc = x \cdot xr \cdot xr^2 = x^3 r^3
\]
Thus, we have:
\[
x^3 r^3 = 46656
\]
### Step 4: Simplify the equation
We can rewrite this as:
\[
(xr)^3 = 46656
\]
Taking the cube root of both sides gives:
\[
xr = \sqrt[3]{46656}
\]
Calculating \( \sqrt[3]{46656} \):
\[
\sqrt[3]{46656} = 36
\]
Thus, we have:
\[
xr = 36
\]
### Step 5: Find \( b - a \)
We know that \( b - a = xr - x = x(r - 1) \) is a perfect cube. Let’s denote \( n^3 = x(r - 1) \) for some positive integer \( n \).
### Step 6: Substitute \( r \)
From \( xr = 36 \), we can express \( r \) as:
\[
r = \frac{36}{x}
\]
Now substituting this into \( b - a \):
\[
b - a = x\left(\frac{36}{x} - 1\right) = 36 - x
\]
Thus, we have:
\[
36 - x = n^3
\]
This gives us:
\[
x = 36 - n^3
\]
### Step 7: Find possible values for \( n \)
Since \( x \) must be positive, we have:
\[
36 - n^3 > 0 \implies n^3 < 36
\]
The possible values for \( n \) are \( 1, 2, 3 \).
1. If \( n = 1 \):
\[
x = 36 - 1^3 = 35
\]
Then \( r = \frac{36}{35} \) (not an integer).
2. If \( n = 2 \):
\[
x = 36 - 2^3 = 28
\]
Then \( r = \frac{36}{28} = \frac{9}{7} \) (not an integer).
3. If \( n = 3 \):
\[
x = 36 - 3^3 = 27
\]
Then \( r = \frac{36}{27} = \frac{4}{3} \) (not an integer).
### Step 8: Check integer values
Since \( r \) must be an integer, we can try different integer values for \( x \) that satisfy \( 36 - x = n^3 \).
Trying \( n = 3 \):
\[
x = 36 - 27 = 9
\]
Then:
\[
r = \frac{36}{9} = 4
\]
### Step 9: Find \( a \), \( b \), and \( c \)
Now substituting back:
- \( a = x = 9 \)
- \( b = xr = 9 \cdot 4 = 36 \)
- \( c = xr^2 = 9 \cdot 16 = 144 \)
### Step 10: Calculate \( a + b + c \)
Now we can find:
\[
a + b + c = 9 + 36 + 144 = 189
\]
Thus, the final answer is:
\[
\boxed{189}
\]
