Facorise: `8a^3+125b^3-64c^3+120abc`

Answer

To factorise the expression \( 8a^3 + 125b^3 - 64c^3 + 120abc \), we can follow these steps:

### Step 1: Rearranging the Expression
We can rearrange the expression to group the terms that fit the sum of cubes and the product of three variables:
\[ 8a^3 + 125b^3 - 64c^3 + 120abc \]

### Step 2: Recognizing the Form
Notice that \( 8a^3 \), \( 125b^3 \), and \( -64c^3 \) can be recognized as cubes:
\[ 8a^3 = (2a)^3, \quad 125b^3 = (5b)^3, \quad -64c^3 = -(4c)^3 \]
Thus, we can rewrite the expression as:
\[ (2a)^3 + (5b)^3 - (4c)^3 + 120abc \]

### Step 3: Applying the Formula
We can apply the formula for the sum of cubes:
\[ x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - xz - yz) \]
Here, let:
- \( x = 2a \)
- \( y = 5b \)
- \( z = -4c \)

### Step 4: Finding \( x + y + z \)
Calculating \( x + y + z \):
\[ x + y + z = 2a + 5b - 4c \]

### Step 5: Finding \( x^2 + y^2 + z^2 \)
Calculating \( x^2 + y^2 + z^2 \):
\[ x^2 = (2a)^2 = 4a^2, \quad y^2 = (5b)^2 = 25b^2, \quad z^2 = (-4c)^2 = 16c^2 \]
Thus,
\[ x^2 + y^2 + z^2 = 4a^2 + 25b^2 + 16c^2 \]

### Step 6: Finding \( xy + xz + yz \)
Calculating \( xy + xz + yz \):
\[ xy = (2a)(5b) = 10ab, \quad xz = (2a)(-4c) = -8ac, \quad yz = (5b)(-4c) = -20bc \]
Thus,
\[ xy + xz + yz = 10ab - 8ac - 20bc \]

### Step 7: Putting It All Together
Now we can substitute back into the formula:
\[ (2a + 5b - 4c)(4a^2 + 25b^2 + 16c^2 - 10ab + 8ac + 20bc) \]

### Final Factorised Form
The final factorised form of the expression is:
\[ (2a + 5b - 4c)(4a^2 + 25b^2 + 16c^2 - 10ab + 8ac + 20bc) \]

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