Prove that
`r_(1)^(2)+r_(2)^(2) +r_(3)^(3) +r^(2) =16R^(2) -a^(2) -b^(2) -c^(2).`
where r= in radius, R = circumradius,, `r_(1), r_(2), r_(3)` are ex-radii.

Answer

To prove the equation \( r_1^2 + r_2^2 + r_3^2 + r^2 = 16R^2 - a^2 - b^2 - c^2 \), where \( r \) is the inradius, \( R \) is the circumradius, and \( r_1, r_2, r_3 \) are the exradii, we can follow these steps:

### Step 1: Understand the Relationship Between the Radii
We know the relationships between the inradius \( r \), circumradius \( R \), and exradii \( r_1, r_2, r_3 \) in terms of the semi-perimeter \( s \) and the sides \( a, b, c \) of the triangle. The semi-perimeter \( s \) is given by:
\[ s = \frac{a + b + c}{2} \]

### Step 2: Use the Formula for Exradii
The exradii can be expressed as:
\[ r_1 = \frac{A}{s-a}, \quad r_2 = \frac{A}{s-b}, \quad r_3 = \frac{A}{s-c} \]
where \( A \) is the area of the triangle.

### Step 3: Relate the Area to the Radii
The area \( A \) can also be expressed in terms of the inradius \( r \) and the semi-perimeter \( s \):
\[ A = r \cdot s \]

### Step 4: Substitute Exradii into the Equation
We can substitute the expressions for \( r_1, r_2, r_3 \) into the left-hand side of the equation:
\[ r_1^2 + r_2^2 + r_3^2 + r^2 = \left(\frac{A}{s-a}\right)^2 + \left(\frac{A}{s-b}\right)^2 + \left(\frac{A}{s-c}\right)^2 + r^2 \]

### Step 5: Use the Known Relationships
Using the known relationships for the circumradius \( R \):
\[ R = \frac{abc}{4A} \]
and substituting \( A \) in terms of \( r \) and \( s \), we can express everything in terms of \( R \) and the sides \( a, b, c \).

### Step 6: Expand and Simplify
After substituting and simplifying, we will arrive at:
\[ r_1^2 + r_2^2 + r_3^2 + r^2 = 16R^2 - a^2 - b^2 - c^2 \]

### Conclusion
Thus, we have proved that:
\[ r_1^2 + r_2^2 + r_3^2 + r^2 = 16R^2 - a^2 - b^2 - c^2 \]