Find the product `(2a + 3b - 4c) (4a ^(2) + 9b ^(2) + 16 c ^(2) - 6 ab + 12 bc +8 ca)`

Answer

To find the product \((2a + 3b - 4c)(4a^2 + 9b^2 + 16c^2 - 6ab + 12bc + 8ca)\), we can use the identity for the product of a sum and a sum of squares.

### Step-by-Step Solution:

**Step 1: Identify the terms.**
We have two expressions:
- First expression: \(2a + 3b - 4c\)
- Second expression: \(4a^2 + 9b^2 + 16c^2 - 6ab + 12bc + 8ca\)

**Hint:** Recognize that the second expression can be rewritten in terms of squares and products.

**Step 2: Rewrite the second expression.**
Notice that:
- \(4a^2 = (2a)^2\)
- \(9b^2 = (3b)^2\)
- \(16c^2 = (-4c)^2\)
- \(-6ab = -2(2a)(3b)\)
- \(12bc = 2(3b)(-4c)\)
- \(8ca = 2(2a)(-4c)\)

This suggests that the second expression can be represented as:
\[ (2a + 3b - 4c)^2 \]

**Hint:** Look for patterns in the coefficients to recognize squares and products.

**Step 3: Apply the identity.**
Using the identity:
\[ (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + xz + yz) \]
we can see that:
\[ (2a + 3b - 4c)(4a^2 + 9b^2 + 16c^2 - 6ab + 12bc + 8ca) = (2a + 3b - 4c)^3 \]

**Hint:** Recognize that the product can be simplified to a cube.

**Step 4: Expand the cube.**
Now we need to expand \((2a + 3b - 4c)^3\):
\[ (2a + 3b - 4c)^3 = (2a)^3 + (3b)^3 + (-4c)^3 + 3(2a)(3b)(-4c) \]
Calculating each term:
- \((2a)^3 = 8a^3\)
- \((3b)^3 = 27b^3\)
- \((-4c)^3 = -64c^3\)
- \(3(2a)(3b)(-4c) = -72abc\)

**Hint:** Use the formula for the cube of a binomial to simplify calculations.

**Step 5: Combine the results.**
Putting it all together:
\[ 8a^3 + 27b^3 - 64c^3 - 72abc \]

**Final Answer:**
The product \((2a + 3b - 4c)(4a^2 + 9b^2 + 16c^2 - 6ab + 12bc + 8ca)\) simplifies to:
\[ 8a^3 + 27b^3 - 64c^3 - 72abc \]