Problem 101 - Algebra and Trigonometry

Question

The Depressed Cubic The most general cubic (third-degree) equation with rational coefficients can be written as $$x^{3}+a x^{2}+b x+c=0$$ (a) Show that if we replace \(x\) by \(X-a / 3\) and simplify, we end up with an equation that doesn't have an \(X^{2}\) term, that is, an equation of the form $$ X^{3}+p X+q=0 $$ This is called a depressed cubic, because we have depressed the quadratic term. (b) Use the procedure described in part (a) to depress the equation \(x^{3}+6 x^{2}+9 x+4=0\)

Step-by-Step Solution

Answer

The depressed cubic is \(X^3 - 3X - 6 = 0\).

Step 1: Express Substitution

To eliminate the quadratic term, we will substitute \( x = X - \frac{a}{3} \) into the original equation, where \( a \) is the coefficient of \( x^2 \).

Step 2: Substitute and Expand

Start with the substitution \( x = X - \frac{a}{3} \). Substitute this into the given cubic equation: \[ (X - \frac{a}{3})^3 + a(X - \frac{a}{3})^2 + b(X - \frac{a}{3}) + c = 0 \] This leads to calculations involving expanding the powers and combining like terms.

Step 3: Expand the Cubic Term

Expand \( (X - \frac{a}{3})^3 \):\[ = X^3 - 3 \cdot \frac{a}{3} X^2 + 3 \cdot \left( \frac{a}{3} \right)^2 X - \left( \frac{a}{3} \right)^3 \]This simplifies to:\[ X^3 - aX^2 + \frac{a^2}{3}X - \frac{a^3}{27} \]

Step 4: Expand the Square Term

Expand \( a(X - \frac{a}{3})^2 \):\[ = a(X^2 - 2 \cdot \frac{a}{3}X + \left( \frac{a}{3} \right)^2) \] This simplifies to:\[ = aX^2 - \frac{2a^2}{3}X + \frac{a^3}{9} \]

Step 5: Combine and Simplify Terms

Combining all terms from the expansion:\[ X^3 - aX^2 + \frac{a^2}{3}X - \frac{a^3}{27} + aX^2 - \frac{2a^2}{3}X + \frac{a^3}{9} + bX - \frac{ab}{3} + c = 0 \]The \(X^2\) terms cancel each other out:\[ X^3 + (b - \frac{2a^2}{3} + \frac{a^2}{3})X + (c - \frac{ab}{3} + \frac{a^3}{9} - \frac{a^3}{27}) = 0 \]

Step 6: Simplify Further

Further simplifying, combine the constant terms:\[ X^3 + (b - \frac{a^2}{3})X + (c - \frac{ab}{3} + \frac{2a^3}{27}) = 0 \] This equation doesn't have the \(X^2\) term anymore.

Step 7: Apply to Given Equation

Now apply this to the specific equation \( x^3 + 6x^2 + 9x + 4 = 0 \).Set \( a = 6 \), \( b = 9 \), \( c = 4 \).Substitute into the formula to recalculate coefficients for the depressed cubic.

Step 8: Calculate New Constants

For \( p \) in the equation:\[ p = 9 - \frac{6^2}{3} = 9 - 12 = -3 \]For \( q \):\[ q = 4 - \frac{6 \times 9}{3} + \frac{2 \times 6^3}{27} = 4 - 18 + 8 = -6 \]

Step 9: Write the Depressed Cubic Equation

From the calculations, the depressed cubic equation is \[ X^3 - 3X - 6 = 0 \] This equation does not have the \(X^2\) term, confirming it is depressed.

Key Concepts

Polynomial Equations

Polynomial equations are expressions involving a sum of powers of a variable, multiplied by coefficients. These equations take the form:

• Linear (degree 1): e.g., \( ax + b = 0 \)
• Quadratic (degree 2): e.g., \( ax^2 + bx + c = 0 \)
• Cubic (degree 3): e.g., \( ax^3 + bx^2 + cx + d = 0 \)

Cubic equations, which we focus on here, are composed of a cubic (\( x^3 \)), a quadratic ( \( x^2 \) ), a linear ( \( x \) ), and a constant term.
If the leading coefficient isn't zero, the highest power determines the equation's degree.
This degree indicates the maximum number of solutions or roots the equation can have. Polynomial equations are fundamental in algebra, used to model a wide range of real-world phenomena.

Substitution Method

The substitution method is a powerful tool in algebra that involves replacing one part of the equation to simplify it.
When applied to cubic equations, it can help in transforming the original equation into a simpler form.
The typical substitution for a cubic equation is:

• Replace \( x \) by \( X - \frac{a}{3} \), where \( a \) is the coefficient of \( x^2 \).

By performing this substitution, we reduce the cubic equation into a depressed cubic,
which simplifies the process of finding solutions. This technique makes use of algebraic manipulation to cancel out terms or reduce complexity,
and is particularly useful in mathematical problem-solving because it can transform a complex polynomial equation into a more manageable form.

Cubic Equation Transformation

Transforming a cubic equation to a depressed cubic is a key step in solving them.
This transformation eliminates the \( X^2 \) term, simplifying the equation to the form \( X^3 + pX + q = 0 \).
How does this work? Using our substitution method,
we set \( x = X - \frac{a}{3} \) and substitute into the original cubic equation:

• First, expand \( (X - \frac{a}{3})^3 \) using binomial expansion.
• Next, expand \( a(X - \frac{a}{3})^2 \).
• Combine these with the linear and constant terms.

Once expanded, reorganize and simplify the equation.
The \( X^2 \) terms cancel out through this process, leaving us with a depressed cubic that is easier to solve.
This method is effective due to its systematic elimination of terms, allowing for focus on the roots of the cubic term.