If `A+B+C=180^(@), (sin 2A+sin 2B+sin2C)/(sinA+sinB+sinC)=k sin. (A)/(2) sin. (B)/(2) sin. (C)/(2)` then the value of `3k^(3)+2k^(2)+k+1` is equal to

Answer

To solve the problem, we start with the given equation:

\[ \frac{\sin 2A + \sin 2B + \sin 2C}{\sin A + \sin B + \sin C} = k \cdot \frac{\sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)}{1} \]

Given that \( A + B + C = 180^\circ \), we can use the property of angles in a triangle. Let’s assume \( A = B = C = 60^\circ \) (an equilateral triangle).

### Step 1: Calculate \( \sin 2A + \sin 2B + \sin 2C \)

Since \( A = B = C = 60^\circ \):

\[ \sin 2A = \sin 120^\circ = \frac{\sqrt{3}}{2} \]
\[ \sin 2B = \sin 120^\circ = \frac{\sqrt{3}}{2} \]
\[ \sin 2C = \sin 120^\circ = \frac{\sqrt{3}}{2} \]

Adding these together:

\[ \sin 2A + \sin 2B + \sin 2C = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \]

### Step 2: Calculate \( \sin A + \sin B + \sin C \)

Again, since \( A = B = C = 60^\circ \):

\[ \sin A = \sin 60^\circ = \frac{\sqrt{3}}{2} \]
\[ \sin B = \sin 60^\circ = \frac{\sqrt{3}}{2} \]
\[ \sin C = \sin 60^\circ = \frac{\sqrt{3}}{2} \]

Adding these together:

\[ \sin A + \sin B + \sin C = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2} \]

### Step 3: Substitute into the equation

Now substituting these results into the original equation:

\[ \frac{\frac{3\sqrt{3}}{2}}{\frac{3\sqrt{3}}{2}} = k \cdot \frac{\sin \left(\frac{60^\circ}{2}\right) \sin \left(\frac{60^\circ}{2}\right) \sin \left(\frac{60^\circ}{2}\right)}{1} \]

This simplifies to:

\[ 1 = k \cdot \left(\sin 30^\circ\right)^3 \]

Since \( \sin 30^\circ = \frac{1}{2} \):

\[ 1 = k \cdot \left(\frac{1}{2}\right)^3 = k \cdot \frac{1}{8} \]

### Step 4: Solve for \( k \)

Multiplying both sides by 8:

\[ k = 8 \]

### Step 5: Calculate \( 3k^3 + 2k^2 + k + 1 \)

Now substituting \( k = 8 \) into the expression:

\[ 3k^3 + 2k^2 + k + 1 = 3(8^3) + 2(8^2) + 8 + 1 \]

Calculating each term:

\[ 8^3 = 512 \quad \Rightarrow \quad 3(512) = 1536 \]
\[ 8^2 = 64 \quad \Rightarrow \quad 2(64) = 128 \]
\[ k = 8 \]
\[ 1 = 1 \]

Adding these together:

\[ 1536 + 128 + 8 + 1 = 1673 \]

### Final Answer

Thus, the value of \( 3k^3 + 2k^2 + k + 1 \) is:

\[ \boxed{1673} \]