If `int(x^(9)+x^(6)+x^(3))(2x^(6)+3x^(3)+6)^(1//3)dx=a(2x^(9)+3x^(6)+6x^(3))^(4//3)+c` then the value of 48a must be _______

Answer

To solve the integral

\[ \int (x^9 + x^6 + x^3)(2x^6 + 3x^3 + 6)^{1/3} \, dx = a(2x^9 + 3x^6 + 6x^3)^{4/3} + c, \]

we will follow a systematic approach.

### Step 1: Factor out \(x^3\) from the integrand
We can factor out \(x^3\) from the polynomial \(x^9 + x^6 + x^3\):

\[ x^9 + x^6 + x^3 = x^3(x^6 + x^3 + 1). \]

Thus, the integral becomes:

\[ \int x^3(x^6 + x^3 + 1)(2x^6 + 3x^3 + 6)^{1/3} \, dx. \]

### Step 2: Substitute \(T\)
Let \(T = 2x^9 + 3x^6 + 6x^3\). We will differentiate \(T\) with respect to \(x\):

\[ \frac{dT}{dx} = 18x^8 + 18x^5 + 18x^2 = 18(x^8 + x^5 + x^2). \]

This gives us:

\[ dT = 18(x^8 + x^5 + x^2) \, dx. \]

### Step 3: Solve for \(dx\)
From the above, we can express \(dx\) in terms of \(dT\):

\[ dx = \frac{dT}{18(x^8 + x^5 + x^2)}. \]

### Step 4: Substitute into the integral
Now we substitute \(T\) and \(dx\) into the integral:

\[ \int x^3(x^6 + x^3 + 1)(2x^6 + 3x^3 + 6)^{1/3} \cdot \frac{dT}{18(x^8 + x^5 + x^2)}. \]

### Step 5: Simplify the integral
Notice that \(x^3(x^6 + x^3 + 1)\) can be simplified. We can express the integral as:

\[ \frac{1}{18} \int T^{1/3} \, dT. \]

### Step 6: Integrate
The integral of \(T^{1/3}\) is:

\[ \int T^{1/3} \, dT = \frac{T^{4/3}}{4/3} = \frac{3}{4} T^{4/3}. \]

Thus, we have:

\[ \frac{1}{18} \cdot \frac{3}{4} T^{4/3} + C = \frac{1}{24} T^{4/3} + C. \]

### Step 7: Substitute back for \(T\)
Substituting back for \(T\):

\[ \frac{1}{24} (2x^9 + 3x^6 + 6x^3)^{4/3} + C. \]

### Step 8: Compare coefficients
Comparing this with the original expression \(a(2x^9 + 3x^6 + 6x^3)^{4/3} + c\), we find:

\[ a = \frac{1}{24}. \]

### Step 9: Calculate \(48a\)
Finally, we calculate \(48a\):

\[ 48a = 48 \cdot \frac{1}{24} = 2. \]

Thus, the value of \(48a\) is

\[ \boxed{2}. \]