Introduction

The problem at hand is to determine the polynomial equation whose roots are the squares of the roots of the given cubic equation ( x^3 - 8 0 ). This involves understanding and applying the principles of polynomial theory and Vieta's formulas.

Step-by-Step Solution

Let's start by finding the roots of the given polynomial equation ( x^3 - 8 0 ).

Step 1: Finding the Roots of the Original Polynomial

The equation ( x^3 - 8 0 ) can be rewritten as:

[ x^3 - 2^3 0 ]

This can be factored using the sum of cubes formula:

[ (x - 2)(x^2 2x 4) 0 ]

Solving for the roots, we get:

α -2 β 1 i√3 γ 1 - i√3

Step 2: Calculating the Squared Roots

We need to find the squares of these roots:

α^2 (-2)^2 4 β^2 (1 i√3)^2 1 2i√3 - 3 -2 2i√3 γ^2 (1 - i√3)^2 1 - 2i√3 - 3 -2 - 2i√3

Step 3: Using Vieta's Formulas

Vieta's formulas relate the coefficients of a polynomial to sums and products of its roots. Given the original polynomial ( x^3 ^2 - 8 0 ), we have:

Sum of the roots ( α β γ 0 ) Sum of the product of the roots taken two at a time ( αβ βγ γα 0 ) Product of the roots ( αβγ -8 )

For the new polynomial with roots ( α^2, β^2, γ^2 ), we need to find the corresponding sums:

Sum of the roots ( α^2 β^2 γ^2 4 (-2 2i√3) (-2 - 2i√3) 0 ) Sum of the product of the roots taken two at a time ( α^2β^2 β^2γ^2 γ^2α^2 (-2 2i√3)(-2 - 2i√3) (-2 2i√3)(4) (4)(-2 - 2i√3) ) Product of the roots ( α^2β^2γ^2 4(-2 2i√3)(-2 - 2i√3) 4(4 12) 64 )

Since ( α^2β^2 β^2γ^2 γ^2α^2 (-2 2i√3)(-2 - 2i√3) (-2 2i√3)(4) (4)(-2 - 2i√3) 4 4 - 8 - 8i√3 8 8i√3 0 )

Step 4: Forming the New Polynomial

The polynomial with roots ( α^2, β^2, γ^2 ) can be written as:

[ x^3 - (α^2 β^2 γ^2)x^2 (α^2β^2 β^2γ^2 γ^2α^2)x - α^2β^2γ^2 0 ]

Substituting the calculated values, we get:

[ x^3 - ^2 - 64 0 ]

This simplifies to:

[ x^3 - 64 0 ]

Conclusion

Therefore, the polynomial whose roots are the squares of the roots of ( x^3 - 8 0 ) is:

[ boxed{x^3 - 64 0} ]

Related Keywords

roots of cubic equations polynomial with given roots Vieta's formulas