If `(x^(2)-yz)/(a) =(y^(2)-zx)/(b) =(z^(2)-xy)/(c )` , then `(x+y+z) (a+b+c) =ax +by +cz`

Answer

To prove the equation \((x+y+z)(a+b+c) = ax + by + cz\) given the condition \(\frac{x^2 - yz}{a} = \frac{y^2 - zx}{b} = \frac{z^2 - xy}{c}\), we will follow these steps:

### Step 1: Set the common ratio
Let us denote the common ratio by \( r \):
\[ \frac{x^2 - yz}{a} = r, \quad \frac{y^2 - zx}{b} = r, \quad \frac{z^2 - xy}{c} = r \]

### Step 2: Rewrite the equations
From the above, we can express each equation in terms of \( r \):
\[ x^2 - yz = ar \quad (1) \]
\[ y^2 - zx = br \quad (2) \]
\[ z^2 - xy = cr \quad (3) \]

### Step 3: Rearranging the equations
Rearranging equations (1), (2), and (3) gives us:
\[ x^2 = ar + yz \quad (4) \]
\[ y^2 = br + zx \quad (5) \]
\[ z^2 = cr + xy \quad (6) \]

### Step 4: Add the equations
Now, we will add equations (4), (5), and (6):
\[ x^2 + y^2 + z^2 = (ar + yz) + (br + zx) + (cr + xy) \]
This simplifies to:
\[ x^2 + y^2 + z^2 = r(a + b + c) + (yz + zx + xy) \quad (7) \]

### Step 5: Using the identity for cubes
Recall the identity:
\[ x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz - zx) \]
From equation (7), we can substitute \( x^2 + y^2 + z^2 \):
\[ x^3 + y^3 + z^3 - 3xyz = (x+y+z)\left(r(a + b + c) + (yz + zx + xy) - xy - yz - zx\right) \]
This simplifies to:
\[ x^3 + y^3 + z^3 - 3xyz = (x+y+z)(ar + br + cr) \]

### Step 6: Factor out \( r \)
We can factor \( r \) out:
\[ x^3 + y^3 + z^3 - 3xyz = r(x+y+z)(a + b + c) \]

### Step 7: Equate both sides
Now, we know from our earlier steps that:
\[ ax + by + cz = r(a + b + c) \]
Thus, we can equate:
\[ (x+y+z)(a+b+c) = ax + by + cz \]

### Conclusion
Thus, we have proved that:
\[ (x+y+z)(a+b+c) = ax + by + cz \]