The equation whose roots are cubes of the roots `x^3 +2x^2 +3=0` is

Options

A. `x^3 +6x^2 -36 x+ 27 =0`
B. `x^3 -5x^3 -11 x +3=0`
C. `x^3 -23 x^2 + 13 x -36 =0`
D. `x^3 +17 x^2 + 27 x+27 =0`

Answer

To find the equation whose roots are the cubes of the roots of the polynomial \( x^3 + 2x^2 + 3 = 0 \), we can follow these steps:

### Step 1: Identify the original polynomial
The original polynomial is given as:
\[ x^3 + 2x^2 + 3 = 0 \]

### Step 2: Let the roots of the original polynomial be \( r_1, r_2, r_3 \)
According to Vieta's formulas, we can express the sums and products of the roots in terms of the coefficients:
- The sum of the roots \( r_1 + r_2 + r_3 = -\frac{b}{a} = -\frac{2}{1} = -2 \)
- The sum of the products of the roots taken two at a time \( r_1r_2 + r_2r_3 + r_3r_1 = \frac{c}{a} = \frac{0}{1} = 0 \)
- The product of the roots \( r_1r_2r_3 = -\frac{d}{a} = -\frac{3}{1} = -3 \)

### Step 3: Find the new roots, which are the cubes of the original roots
We need to find the new roots \( s_1, s_2, s_3 \) where:
\[ s_i = r_i^3 \quad (i = 1, 2, 3) \]

### Step 4: Use the relationships between the roots
Using the relationships from Vieta's formulas for the new roots:
- The sum of the new roots:
\[ s_1 + s_2 + s_3 = r_1^3 + r_2^3 + r_3^3 \]
Using the identity \( r_1^3 + r_2^3 + r_3^3 = (r_1 + r_2 + r_3)(r_1^2 + r_2^2 + r_3^2 - r_1r_2 - r_2r_3 - r_3r_1) + 3r_1r_2r_3 \):
- First, we find \( r_1^2 + r_2^2 + r_3^2 \):
\[ r_1^2 + r_2^2 + r_3^2 = (r_1 + r_2 + r_3)^2 - 2(r_1r_2 + r_2r_3 + r_3r_1) = (-2)^2 - 2(0) = 4 \]
Thus,
\[ r_1^3 + r_2^3 + r_3^3 = (-2)(4 - 0) + 3(-3) = -8 - 9 = -17 \]

### Step 5: Find the sum of the products of the new roots
The sum of the products of the new roots taken two at a time:
\[ s_1s_2 + s_2s_3 + s_3s_1 = r_1^3r_2^3 + r_2^3r_3^3 + r_3^3r_1^3 = (r_1r_2)^3 + (r_2r_3)^3 + (r_3r_1)^3 \]
Using the identity \( (a + b + c)^3 = a^3 + b^3 + c^3 + 3(a + b)(b + c)(c + a) \):
\[ = 0^3 - 3(-3) = 9 \]

### Step 6: Find the product of the new roots
The product of the new roots:
\[ s_1s_2s_3 = (r_1r_2r_3)^3 = (-3)^3 = -27 \]

### Step 7: Form the new polynomial
Using Vieta's relations for the new roots, we can write the polynomial:
\[ y^3 - (s_1 + s_2 + s_3)y^2 + (s_1s_2 + s_2s_3 + s_3s_1)y - s_1s_2s_3 = 0 \]
Substituting the values we found:
\[ y^3 + 17y^2 + 9y + 27 = 0 \]

### Step 8: Substitute back to the variable \( x \)
Since we let \( y = x \), the polynomial becomes:
\[ x^3 + 17x^2 + 9x + 27 = 0 \]

Thus, the equation whose roots are the cubes of the roots of \( x^3 + 2x^2 + 3 = 0 \) is:
\[ \boxed{x^3 + 17x^2 + 9x + 27 = 0} \]