In `DeltaABC, angle C = 2 angle A, and AC = 2BC`, then the value of `(a^(2) + b^(2) c^(2))/(R^(2))` (where R is circumradius of triangle) is ______

Explanation

To solve the problem, we need to find the value of \((a^2 + b^2 + c^2) / R^2\) given that in triangle \(ABC\), \(\angle C = 2\angle A\) and \(AC = 2BC\).

### Step-by-Step Solution:

1. **Set up the angles**:
Let \(\angle A = A\), then \(\angle C = 2A\). The sum of angles in a triangle gives us:
\[ A + B + C = 180^\circ \]
Substituting for \(C\):
\[ A + B + 2A = 180^\circ \implies 3A + B = 180^\circ \implies B = 180^\circ - 3A \]

2. **Use the side length condition**:
We are given that \(AC = 2BC\). In terms of sides, we denote:
- \(a = BC\)
- \(b = AC\)
- \(c = AB\)

From the condition \(AC = 2BC\), we have:
\[ b = 2a \]

3. **Apply the Sine Rule**:
According to the sine rule:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
Substituting \(b = 2a\):
\[ \frac{a}{\sin A} = \frac{2a}{\sin B} \]
This simplifies to:
\[ \sin B = 2 \sin A \]

4. **Express \(\sin B\) in terms of \(\sin A\)**:
Using the angle \(B = 180^\circ - 3A\):
\[ \sin B = \sin(180^\circ - 3A) = \sin 3A \]
Therefore, we have:
\[ \sin 3A = 2 \sin A \]

5. **Use the sine triple angle formula**:
The sine triple angle formula states:
\[ \sin 3A = 3 \sin A - 4 \sin^3 A \]
Setting this equal to \(2 \sin A\):
\[ 3 \sin A - 4 \sin^3 A = 2 \sin A \]
Rearranging gives:
\[ \sin A - 4 \sin^3 A = 0 \]
Factoring out \(\sin A\):
\[ \sin A (1 - 4 \sin^2 A) = 0 \]
This gives us:
- \(\sin A = 0\) (not valid for a triangle)
- \(1 - 4 \sin^2 A = 0 \implies \sin^2 A = \frac{1}{4} \implies \sin A = \frac{1}{2}\)

6. **Find angles**:
If \(\sin A = \frac{1}{2}\), then:
\[ A = 30^\circ \quad (\text{or } \frac{\pi}{6}) \]
Thus,
\[ C = 2A = 60^\circ \quad (\text{or } \frac{\pi}{3}) \]
And,
\[ B = 180^\circ - 3A = 90^\circ \quad (\text{or } \frac{\pi}{2}) \]

7. **Calculate sides using circumradius**:
The circumradius \(R\) relates to the sides as follows:
\[ a = 2R \sin A, \quad b = 2R \sin B, \quad c = 2R \sin C \]
Substituting the values:
- \(a = 2R \cdot \frac{1}{2} = R\)
- \(b = 2R \cdot 1 = 2R\)
- \(c = 2R \cdot \frac{\sqrt{3}}{2} = R\sqrt{3}\)

8. **Calculate \(a^2 + b^2 + c^2\)**:
\[ a^2 + b^2 + c^2 = R^2 + (2R)^2 + (R\sqrt{3})^2 = R^2 + 4R^2 + 3R^2 = 8R^2 \]

9. **Find \(\frac{a^2 + b^2 + c^2}{R^2}\)**:
\[ \frac{a^2 + b^2 + c^2}{R^2} = \frac{8R^2}{R^2} = 8 \]

### Final Answer:
The value of \(\frac{a^2 + b^2 + c^2}{R^2}\) is \(8\).