Posted by **candice** on Monday, September 11, 2006 at 9:17pm.

The roots of the eqn, x^4 + px^3 + qx^2 + rx + s = 0 where p, q, r, s are constants and s does not equal to 0, are a, b, c, d.

(i) a^2 + b^2 + c^2 + d^2 = p^2 -2q (in terms of p & q)

(ii) 1/a + 1/b + 1/c + 1/d = -r/s

(in terms of r & s)

(iii) using the above results, or otherwise, express a^3 + b^3 + c^3 + d^3 in terms of p, q & r.

This looks like modern algebra, not high school algebra. I'm not sure how much time I'll devote to this.

You should know that p is sigma-1(roots) and q is sigma-2(roots). If you know how to work with the symmetric polynomials then the result follows immediately for (i).

If you add the terms in (ii) you get -sigma-3(roots) / sigma-4(roots) and the result follows immediately

For the last part a^3 + b^3 + c^3 + d^3 =

(sigma-1)^3-3(sigma-1)(sigma-2)+3(sigma-3) all of them of the roots

p=-sigma-1, q=sigma-2, r=sigma-3 so (iii) is

(-p)^3-3(-p)q+3r=a^3+b^3+c^3+d^3

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