Transform the following equation into one in which the coefficients of the third highest power of x is zero . `x^3+2x^2+x+1=0`

Answer

To transform the equation \( x^3 + 2x^2 + x + 1 = 0 \) into one where the coefficient of the third highest power of \( x \) is zero, we will follow these steps:

### Step 1: Substitute \( x \) with \( y + a \)
Let \( x = y + a \). This substitution will help us manipulate the equation to eliminate the \( y^2 \) term.

### Step 2: Expand the equation
Substituting \( x \) into the equation gives:
\[ (y + a)^3 + 2(y + a)^2 + (y + a) + 1 = 0 \]
Now, we expand each term:
- \( (y + a)^3 = y^3 + 3ay^2 + 3a^2y + a^3 \)
- \( 2(y + a)^2 = 2(y^2 + 2ay + a^2) = 2y^2 + 4ay + 2a^2 \)
- \( (y + a) = y + a \)

Combining these, we have:
\[ y^3 + 3ay^2 + 3a^2y + a^3 + 2y^2 + 4ay + 2a^2 + y + a + 1 = 0 \]

### Step 3: Combine like terms
Now, we collect the coefficients of like powers of \( y \):
- Coefficient of \( y^3 \): \( 1 \)
- Coefficient of \( y^2 \): \( 3a + 2 \)
- Coefficient of \( y \): \( 3a^2 + 4a + 1 \)
- Constant term: \( a^3 + 2a^2 + a + 1 \)

Thus, we can rewrite the equation as:
\[ y^3 + (3a + 2)y^2 + (3a^2 + 4a + 1)y + (a^3 + 2a^2 + a + 1) = 0 \]

### Step 4: Set the coefficient of \( y^2 \) to zero
To eliminate the \( y^2 \) term, we set:
\[ 3a + 2 = 0 \]
Solving for \( a \):
\[ 3a = -2 \implies a = -\frac{2}{3} \]

### Step 5: Substitute \( a \) back into the equation
Now, substitute \( a = -\frac{2}{3} \) back into the coefficients of \( y \):
- Coefficient of \( y \):
\[ 3\left(-\frac{2}{3}\right)^2 + 4\left(-\frac{2}{3}\right) + 1 = 3 \cdot \frac{4}{9} - \frac{8}{3} + 1 = \frac{4}{3} - \frac{8}{3} + 1 = -\frac{4}{3} + 1 = -\frac{1}{3} \]

- Constant term:
\[ \left(-\frac{2}{3}\right)^3 + 2\left(-\frac{2}{3}\right)^2 + \left(-\frac{2}{3}\right) + 1 = -\frac{8}{27} + 2 \cdot \frac{4}{9} - \frac{2}{3} + 1 \]
Converting to a common denominator (27):
\[ -\frac{8}{27} + \frac{24}{27} - \frac{18}{27} + \frac{27}{27} = \frac{25 - 26}{27} = -\frac{1}{27} \]

### Final equation
Thus, the transformed equation is:
\[ y^3 - \frac{1}{3}y - \frac{1}{27} = 0 \]

### Summary
The final transformed equation does not contain the \( y^2 \) term, effectively making the coefficient of the third highest power of \( x \) zero.

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