Let α, β, γ be the three roots of the equation x³+b x+c=0. If β γ=1=-α, then b³+2 c³-3 α³-6 β³-8 γ³ is equal to :

Solution

Maths sir · 4 steps

Get b and c from Vieta, then notice that β and γ are cube roots of −1, so every cube is easy.

1. Use Vieta. There is no x² term, so α + β + γ = 0. With α = -1, this gives β + γ = 1. Also αβγ = -c, so (-1)(1) = -c and c = 1.
2. Find b. b = αβ + βγ + γα = α(β + γ) + βγ = (-1)(1) + 1 = 0. Check: x = -1 satisfies x³ + 1 = 0.
3. Cubes of roots. With β + γ = 1 and βγ = 1, the numbers β and γ are the roots of t² - t + 1 = 0. Multiply by (t + 1) to get t³ + 1 = 0, so β³ = γ³ = -1. Also α³ = -1.

   b³ + 2c³ - 3α³ - 6β³ - 8γ³ = 0 + 2 + 3 + 6 + 8

4. So. The expression equals 19, which is option (b).

The idea to keep. When two roots satisfy t² - t + 1 = 0 or t² + t + 1 = 0, multiply by (t+1) or (t-1). You get t³ = -1 or t³ = 1, and every power of those roots becomes easy.

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