^{3}+ px - 1 = 0 has three real, non-zero roots (α,β,γ)

a) Find the values of α

^{2}+ β

^{2}+ γ

^{2}and α

^{4}+ β

^{4}+ γ

^{4}in terms of p, and hence show that p is negative.

b) Find the monic cubic equation with co-efficients in terms of p, whose roots are α

^{2}, β

^{2}, γ

^{2}

I've completed part a), and almost completed part b), but my problem is that the solution is x

^{3}+ 2px

^{2}- 1 = 0

So far, I have managed to construct a cubic with those roots, but I have x

^{3}+ 2px

^{2}+ p

^{2}x - 1 = 0.

I'm not sure if the answer is wrong or not (as per usual)

a) α+β+γ = -b/a = 0

α

^{2}+ β

^{2}+ γ

^{2}= (α+β+γ)

^{2}- 2(αβ+βγ+γα) = -2p

α

^{2}+ β

^{2}+ γ

^{2}> 0

Sum of squares of real non-zeroes must be strictly positive

-2p > 0

p < 0

Not sure how sum of fourth powers helps, since this can be deduced from sum of squares.

b) let y = x

^{2}and hence, we will deduce the new polynomial from substitution

x

^{3}+ px - 1 = 0

x

^{3}+ px = 1

x(x

^{2}+ p) = 1

x

^{2}(x

^{2}+ p)

^{2}= 1

y(y + p)

^{2}= 1

y

^{3}+ 2py

^{2}+ p

^{2}y = 1

y

^{3}+ 2py

^{2}+ p

^{2}y -1 = 0

replacing y with x for the required polynomial with roots α

^{2}, β

^{2}, γ

^{2}

x

^{3}+ 2px

^{2}+ p

^{2}x - 1 = 0