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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 ?​

User Vivek Hirpara
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1 Answer

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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

User Balping
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