Question

brain ly moderators, brainly stars pls answer this question​

Answer 1

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

And very simple solution by @pulakmath007  in attachment :

https://brainly.in/profile/pulakmath007-13087864

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

Learn More:

if alpha beta and gamma are the zeros of the polynomial x cube brainly.in/question/10667886

find the centroid of the triangle xyz whose vertices are x 3, -5, y -3, 4 ...

brainly.in/question/14122475

Answer 2

Given :- The equation x³ - 3x + 1 = 0 has roots a, b and c. Find a cubic equation with integers coefficients that has roots a², b² and c² ? { I am assuming alpha as a , beta as b and gama as c .}

Solution :-

we know that, for the cubic equation P(x) =ax³ + bx² + cx + d = 0 ,a ≠ 0.

• If α, β and γ are zeros of P(x),

Then :-

⟿ α + β + γ = (-b)/a

⟿ αβ + βγ + αγ = c/a

⟿ αβγ = (-d)/a

comparing given cubic equation x³ - 3x + 1 = 0 with ax³ + bx² + cx + d = 0 , we get,

• a = 1
• b = 0
• c = (-3)
• d = 1

then,

→ a + b + c = 0

→ ab + bc + ca = (-3)

→ abc = -1

so,

→ (a + b + c) = a² + b² + c² + 2(ab + bc + ca)

→ 0 = a² + b² + c² + 2 * (-3)

→ a² + b² + c² = 6

and,

→ (abc)² = (-1)²

→ a²b²c² = 1

and,

→ (ab + bc + ca)² = a²b² + b²c² + c²a² + 2(ab²c + bc²a + ca²b) = a²b² + b²c² + c²a² + 2abc(a + b + c)

→ (-3)² = a²b² + b²c² + c²a² + 2 * (-1) * 0

→ a²b² + b²c² + c²a² = 9

comparing again,

→ a² + b² + c² = sum of roots = -b/a = 6

• b = (-6)
• a = 1

and,

→ a²b²c² = product of roots = -d/a = 1

• d = (-1)
• a = 1

and,

→ a²b² + b²c² + ²a² = c/a = 9

• c = 9
• a = 1

therefore, required cubic equation, whose roots are a², b² and c² will be ,

→ ax³ + bx² + cx + d = 0

→ x³ - 6x² + 9x - 1 = 0

Learn more :-

Let a, b and c be non-zero real numbers satisfying (a³)/(b³ + c³) + (b³)/(c³ + a³) + (c³)/(a³ + b³)

https://brainly.in/question/40626097

https://brainly.in/question/20858452

if a²+ab+b²=25

b²+bc+c²=49

c²+ca+a²=64

Then, find the value of

(a+b+c)² - 100 = __

https://brainly.in/question/16231132