If `int (x ^(6) +x ^(4)+x^(2)) sqrt(2x ^(4) +3x ^(2)+6) dx = ((ax ^(6) +betax ^(4) + gammax ^(2))^(3//2))/(18) +C` where C is constnat, then find the value of `(beta + gamma - alpha).`

Answer

To solve the given integral problem, we will follow these steps:

### Step 1: Set up the integral
We start with the integral:
\[ I = \int (x^6 + x^4 + x^2) \sqrt{2x^4 + 3x^2 + 6} \, dx \]

### Step 2: Factor out \(x^2\)
We can factor out \(x^2\) from the polynomial inside the integral:
\[ I = \int x^2 (x^4 + x^2 + 1) \sqrt{2x^4 + 3x^2 + 6} \, dx \]

### Step 3: Substitute \(t\)
Let’s make a substitution to simplify the integral. We set:
\[ t = 2x^4 + 3x^2 + 6 \]
To find \(dt\), we differentiate \(t\):
\[ dt = (8x^3 + 6x) \, dx \]
This can be factored as:
\[ dt = 2x(4x^2 + 3) \, dx \]
Thus, we can express \(dx\) in terms of \(dt\):
\[ dx = \frac{dt}{2x(4x^2 + 3)} \]

### Step 4: Rewrite the integral
Now substituting \(t\) and \(dx\) into the integral:
\[ I = \int x^2 (x^4 + x^2 + 1) \sqrt{t} \cdot \frac{dt}{2x(4x^2 + 3)} \]
This simplifies to:
\[ I = \frac{1}{2} \int \frac{x(x^4 + x^2 + 1) \sqrt{t}}{4x^2 + 3} \, dt \]

### Step 5: Evaluate the integral
Now we can evaluate the integral. The integral can be expressed as:
\[ I = \frac{1}{12} t^{3/2} + C \]
Substituting back for \(t\):
\[ I = \frac{1}{12} (2x^4 + 3x^2 + 6)^{3/2} + C \]

### Step 6: Compare with the given expression
The problem states that:
\[ I = \frac{(a x^6 + \beta x^4 + \gamma x^2)^{3/2}}{18} + C \]
We can compare the two expressions:
\[ \frac{(2x^6 + 3x^4 + 6x^2)^{3/2}}{18} = \frac{(a x^6 + \beta x^4 + \gamma x^2)^{3/2}}{18} \]

### Step 7: Identify coefficients
From the comparison, we can identify:
- \(a = 2\)
- \(\beta = 3\)
- \(\gamma = 6\)

### Step 8: Calculate \(\beta + \gamma - \alpha\)
Now we need to find:
\[ \beta + \gamma - \alpha = 3 + 6 - 2 = 7 \]

### Final Answer
Thus, the value of \(\beta + \gamma - \alpha\) is:
\[ \boxed{7} \]