Answer:
x^3 - 6x^2 + 12x - 8 = 0
Explanation:
Let the roots of the equation x^3 - 8x + 3 = 0 be a, b, and c. Then, we know that the squares of the roots are a^2, b^2, and c^2.
By Vieta's formulas, we know that the sum of the roots of x^3 - 8x + 3 = 0 is 0, since there is no x^2 term:
a + b + c = 0
We also know that the product of the roots is 3/1 = 3:
abc = 3
Using the fact that the squares of the roots are a^2, b^2, and c^2, we can write:
a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc)
Since a + b + c = 0, this simplifies to:
a^2 + b^2 + c^2 = -2(ab + ac + bc)
Now, let's try to find a cubic equation with roots a^2, b^2, and c^2. We can start with:
(x - a^2)(x - b^2)(x - c^2) = 0
Expanding the left side, we get:
x^3 - (a^2 + b^2 + c^2)x^2 + (a^2b^2 + b^2c^2 + c^2a^2)x - a^2b^2c^2 = 0
Using the identity we found earlier for a^2 + b^2 + c^2, we can simplify this to:
x^3 + 2(ab + ac + bc)x - 3a^2b^2c^2 = 0
Now, we can substitute in the values we know for ab, ac, and bc:
ab = (a + b + c)(ab + ac + bc) - (a^2b + ab^2 + c^2a)
ac = (a + b + c)(ab + ac + bc) - (a^2c + ac^2 + b^2a)
bc = (a + b + c)(ab + ac + bc) - (b^2c + bc^2 + a^2b)
Plugging in these values and simplifying, we get:
x^3 - 6x^2 + 12x - 8 = 0
Therefore, the cubic equation we're looking for is:
x^3 - 6x^2 + 12x - 8 = 0
And the roots of this equation are the squares of the roots of x^3 - 8x + 3 = 0.