Question

The sum of the cubes of all the roots of the equation x⁴ – 3x³ –2x² + 3x +1 = 0 is

Answer 1

Correct Answer: 36

The correct answer is 36
x⁴ – 3x³ – x² – x² + 3x + 1 = 0
(x² – 1) (x² – 3x – 1) = 0
Let the root of x² – 3x – 1 = 0 be α and β and other two roots of given equation are 1 and –1
So sum of cubes of roots = 1³ + (–1)³ + α³ + β³
= (α + β)³ – 3αβ(α + β)
= (3)³ – 3(–1)(3)
= 36

Answer 2

Consider the polynomial:

\( x^4 - 3x^3 - 2x^2 + 3x + 1 = 0 \)

This polynomial factors as:

\( (x^2 - 1)(x^2 - 3x - 1) = 0 \)

Thus, its roots are:

• \( x = 1 \) and \( x = -1 \) from \( x^2 - 1 = 0 \)
• \( \alpha \) and \( \beta \) from \( x^2 - 3x - 1 = 0 \)

Step 1: Use Vieta's Formulas

For the quadratic equation \( x^2 - 3x - 1 = 0 \), Vieta's formulas yield:

• Sum of roots: \( \alpha + \beta = 3 \)
• Product of roots: \( \alpha\beta = -1 \)

Step 2: Compute the Sum of Cubes of All Roots

The sum of the cubes of the roots is:

\( 1^3 + (-1)^3 + \alpha^3 + \beta^3 \)

Notice that:

\( 1^3 + (-1)^3 = 1 - 1 = 0 \)

So, we only need to find:

\( \alpha^3 + \beta^3 \)

Using the identity:

\( \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) \)

Substitute the values \( \alpha + \beta = 3 \) and \( \alpha\beta = -1 \):

\( \alpha^3 + \beta^3 = 3^3 - 3(-1)(3) = 27 + 9 = 36 \)

Final Answer

Therefore, the sum of the cubes of all the roots of the given polynomial is 36.