For a cubic poltnomial `a_3x^3+a_2x^2+a_1x+a_0` the roots `r_1,r_2,r_3` come in three symmetric combinations: `C_1=r_1+r_2+r_3,C_2=r_1r_2+r_2r_3+r_3r_1,C_3=r_1r_2r_3` The sum of the `k^(th)` powers of the roots is defined as `S_k=r_1^k+r_2^k+r_3^k` Then
`a_3S_1+a_2=0`
`a_3S_2+a_2S_1+2a_1=0`

Options

A. only (i) is true
B. only (ii) is true
C. both are true
D. none are true

Answer

To solve the problem, we need to analyze the given cubic polynomial and its roots. The polynomial is given as:

\[ a_3x^3 + a_2x^2 + a_1x + a_0 = 0 \]

with roots \( r_1, r_2, r_3 \). The symmetric sums of the roots are defined as:

- \( C_1 = r_1 + r_2 + r_3 \)
- \( C_2 = r_1r_2 + r_2r_3 + r_3r_1 \)
- \( C_3 = r_1r_2r_3 \)

The sum of the \( k^{th} \) powers of the roots is defined as:

\[ S_k = r_1^k + r_2^k + r_3^k \]

We are given two equations to verify:

1. \( a_3 S_1 + a_2 = 0 \)
2. \( a_3 S_2 + a_2 S_1 + 2a_1 = 0 \)

### Step 1: Calculate \( S_1 \)

By Vieta's formulas, we know:

\[ S_1 = r_1 + r_2 + r_3 = -\frac{a_2}{a_3} \]

### Step 2: Substitute \( S_1 \) into the first equation

Substituting \( S_1 \) into the first equation:

\[ a_3 S_1 + a_2 = a_3 \left(-\frac{a_2}{a_3}\right) + a_2 \]

This simplifies to:

\[ -a_2 + a_2 = 0 \]

Thus, the first equation holds true.

### Step 3: Calculate \( S_2 \)

To find \( S_2 \), we use the identity:

\[ S_2 = r_1^2 + r_2^2 + r_3^2 = (r_1 + r_2 + r_3)^2 - 2(r_1r_2 + r_2r_3 + r_3r_1) \]

Substituting the known values:

\[ S_2 = S_1^2 - 2C_2 \]

Substituting \( S_1 = -\frac{a_2}{a_3} \):

\[ S_2 = \left(-\frac{a_2}{a_3}\right)^2 - 2C_2 \]

Now, we know \( C_2 = \frac{a_1}{a_3} \) (from Vieta's formulas), so:

\[ S_2 = \frac{a_2^2}{a_3^2} - 2\left(\frac{a_1}{a_3}\right) \]

### Step 4: Substitute \( S_2 \) and \( S_1 \) into the second equation

Now we substitute \( S_2 \) and \( S_1 \) into the second equation:

\[ a_3 S_2 + a_2 S_1 + 2a_1 = 0 \]

Substituting the values we found:

\[ a_3 \left(\frac{a_2^2}{a_3^2} - 2\frac{a_1}{a_3}\right) + a_2 \left(-\frac{a_2}{a_3}\right) + 2a_1 = 0 \]

This simplifies to:

\[ \frac{a_3 a_2^2}{a_3^2} - 2a_1 + \left(-\frac{a_2^2}{a_3}\right) + 2a_1 = 0 \]

Combining terms gives:

\[ \frac{a_2^2}{a_3} - 2a_1 + 2a_1 = 0 \]

Thus, we find that:

\[ \frac{a_2^2}{a_3} = 0 \]

This confirms that the second equation also holds true.

### Conclusion

Both equations are satisfied, confirming the relationships between the coefficients and the roots of the cubic polynomial.