Let the roots of x³ + 2x + 1 = 0 be a, b, c. The question requires us to find a cubic polynomial whose roots are a², b², c². I can think of two methods to do this. Since the question is posted under MX1, let's start with the method that uses knowledge from 2U/MX1 courses.
Method 1:
Consider x³ + 2x + 1 = 0. Using sums and products of roots:
a + b + c = 0
ab + bc + ca = 2
abc = -1
Let the required cubic polynomial be in the form of px³ + qx² + rx + s = 0. Using sums and products of roots:
a² + b² + c² = -q/p
a²b² + b²c² + c²a² = r/p
a²b²c² = -s/p
However,
a² + b² + c² = (a + b + c)² - 2(ab + bc + ca)
.: -q/p = 0² - 2(2) = -4 --> q = 4p
a²b² + b²c² + c²a² = (ab + bc + ca)²- 2(a²bc + ab²c + abc²) = (ab + bc + ca)² - 2abc(a + b + c)
.: r/p = 2² - 2(-1)(0) = 4 --> r = 4p
a²b²c² = (abc)²
.: -s/p = (-1)² --> s = -p
So the required cubic polynomial can be written as px³ + 4px² + 4px - p = 0.
Dividing by p gives x³ + 4x² + 4x - 1 = 0 as the simplest form.
As you can see, Method 1 can become impossibly tiresome if dealing with polynomials of higher degrees. The next method is the standard MX2 procedure of finding the equation of a polynomial whose roots are related to those of another given polynomial. Even if you don't do MX2, this trick is probably still worthwhile knowing.
Method 2:
Consider x³ + 2x + 1 = 0, with roots a, b, c.
Let y = x². Then clearly P(y) = 0 has roots a², b², c².
y = x² --> x = √y
Substituting this into x³ + 2x + 1 = 0:
(√y)³ + 2√y + 1 = 0
(√y)(y + 2) = -1
[(√y)(y + 2)]² = (-1)²
y(y² + 4y + 4) = 1
.: P(y) = y³ + 4y² + 4y - 1 = 0 has roots a², b², c².
Since x and y are both arbitrary variables, the required cubic polynomial can be expressed in terms of x as x³ + 4x² + 4x - 1 = 0.
