Remove second term ( second higher power of x ) from the equation
` x^3-6x^2 +4x -7=0`

Answer

To solve the equation \( x^3 - 6x^2 + 4x - 7 = 0 \) by removing the second term (the second higher power of \( x \), which is \( -6x^2 \)), we will perform a substitution. Let's go through the steps:

### Step 1: Substitute \( x = y + h \)
We will substitute \( x \) with \( y + h \), where \( h \) is a constant that we will determine later.

### Step 2: Expand the equation
Substituting \( x \) in the equation gives us:
\[ (y + h)^3 - 6(y + h)^2 + 4(y + h) - 7 = 0 \]

### Step 3: Expand \( (y + h)^3 \) and \( (y + h)^2 \)
Using the binomial expansion:
\[ (y + h)^3 = y^3 + 3y^2h + 3yh^2 + h^3 \]
\[ (y + h)^2 = y^2 + 2yh + h^2 \]
Now substituting these expansions into the equation:
\[ y^3 + 3y^2h + 3yh^2 + h^3 - 6(y^2 + 2yh + h^2) + 4(y + h) - 7 = 0 \]

### Step 4: Simplify the equation
Now we simplify the equation:
\[ y^3 + 3y^2h + 3yh^2 + h^3 - 6y^2 - 12yh - 6h^2 + 4y + 4h - 7 = 0 \]

### Step 5: Collect like terms
Collecting the terms based on powers of \( y \):
\[ y^3 + (3h - 6)y^2 + (3h^2 - 12h + 4)y + (h^3 - 6h^2 + 4h - 7) = 0 \]

### Step 6: Remove the \( y^2 \) term
To remove the \( y^2 \) term, we set the coefficient of \( y^2 \) to zero:
\[ 3h - 6 = 0 \]
Solving for \( h \):
\[ 3h = 6 \implies h = 2 \]

### Step 7: Substitute \( h \) back into the equation
Now substitute \( h = 2 \) back into the equation:
\[ y^3 + (3(2)^2 - 12(2) + 4)y + (2^3 - 6(2^2) + 4(2) - 7) = 0 \]
Calculating the coefficients:
- For \( y \): \( 3(2) - 12 + 4 = 6 - 12 + 4 = -2 \)
- For the constant term: \( 8 - 24 + 8 - 7 = -15 \)

### Final Equation
Thus, the equation simplifies to:
\[ y^3 - 2y - 15 = 0 \]

### Summary of Steps:
1. Substitute \( x = y + h \).
2. Expand the equation.
3. Simplify and collect like terms.
4. Set the coefficient of \( y^2 \) to zero to find \( h \).
5. Substitute \( h \) back to get the final equation.