This was a past AIME problem. An answer was as follows:9) The roots of x<sup>4</sup> - x<sup>3</sup> - x<sup>2</sup> - 1 = 0 are a, b, c, and d. Find p(a) + p(b) + p(c) + p(d), where p(x) = x<sup>6</sup> - x<sup>5</sup> - x<sup>3</sup> - x<sup>2</sup> - x.
. . . . .x<sup>4</sup> - x<sup>3</sup> - x<sup>2</sup> - 1 = (x + 1)(x<sup>3</sup> - 2x<sup>2</sup> + x - 1), so:
. . . . . . .a = -1
. . . . . . .b + c + d = 2
. . . . . . .bc + cd + db = 1
. . . . .Hence b<sup>2</sup>+c<sup>2</sup>+d<sup>2</sup> = 2<sup>2</sup> - 2 = 2.
. . . . .We have:
. . . . . . .p(x) = (x<sup>3</sup> - 2x<sup>2</sup> + x - 1)(x<sup>3</sup> + x<sup>2</sup> + x + 1) + x<sup>2</sup> - x + 1
. . . . .Hence:
. . . . . . .p(a) = 3
. . . . . . .p(b) = b<sup>2</sup> - b + 1
. . . . . . .p(c) = c<sup>2</sup> - c + 1
. . . . . . .p(d) = d<sup>2</sup> - d + 1
. . . . .so:
. . . . .p(a) + p(b) + p(c) + p(d)
. . . . . . .= 6 + (b<sup>2</sup> + c<sup>2</sup> + d<sup>2</sup>) - (b + c + d)
. . . . . . .= 6
Can someone please esplicitly explain this to me.