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

Solution

Maths sir · 4 steps

Do not hunt for the roots. Build the sum of cubes from the Vieta sums the equation hands you.

1. Read off Vieta. For x⁴ - 3x³ - 2x² + 3x + 1 = 0 with roots α, β, γ, δ, the signs alternate. Σ α = 3, Σ αβ = -2, and Σ αβγ = -3.
2. Use the identity. The sum of cubes of the roots can be written using only these three sums. This holds for any number of roots.

   Σ α³ = (Σ α)³ - 3(Σ α)(Σ αβ) + 3Σ αβγ

3. Put in numbers. Σ α³ = 3³ - 3(3)(-2) + 3(-3) = 27 + 18 - 9.
4. So. The sum of the cubes of all the roots is 36.

The common slip: 54. Taking Σ αβγ = +3 instead of -3 gives 27 + 18 + 9 = 54. For a quartic, the triple-product sum is -d/a.

The idea to keep. Any symmetric sum of roots, such as squares or cubes, can be built from the Vieta sums. Write the identity first, then fill in the signs carefully.

Answer from the NTA final answer key. Solution by PrepWiser.