Step-by-step explanation:
Given : x³ - 3x + 1 = 0 has roots α , β and γ
To Find : Cubic equation with roots α² , β² and γ²
Solution:
x³ - 3x + 1 = 0 has roots α , β and γ
α + β + γ = 0 as no coefficient of x²
αβ + βγ + αγ = - 3
αβγ = - 1
α² + β² + γ² = (α + β + γ )² - 2(αβ + βγ + αγ)
= 0 - 2(-3)
= 6
α² β² + β² γ² + α² γ² = ?
αβ + βγ + αγ = - 3
Squaring both sides
=>α² β² + β² γ² + α² γ² + 2αβγ( α + β + γ ) = 9
=> α² β² + β² γ² + α² γ² + 0 = 0
=> α² β² + β² γ² + α² γ² = 9
α² β² γ² = (αβγ )² = 1
α² + β² + γ² = 6
α² β² + β² γ² + α² γ² = 9
α² β² γ² = 1
x³ - 6x² + 9x - 1 = 0
x³ - 6x² +9x - 1 = 0 is the required equation
its given α , β , γ are roots and α² , β² , γ² are roots of new
so just replace x with √x in x³ - 3x + 1 = 0
=> (√x)³ - 3√x + 1 = 0
=> x√x - 3√x = - 1
=> √x(x - 3) = - 1
Squaring both sides
=> x(x - 3)² = 1
=> x( x² - 6x + 9) = 1
=> x³ - 6x² + 9x - 1 = 0
