The equation x³-3x+1 = 0 has roots α, β and γ. Find a cubic equation with integer coefficients that has roots α², β² and γ² ?

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

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