50 2 Algebra
see that the numbers 2 cos^27 π, 2 cos^47 π, and 2 cos^87 πare of the formx+^1 x, withxone of
these roots.
If we definey=x+^1 x, thenx^2 +x^12 =y^2 −2 andx^3 +x^13 =y^3 − 3 y. Dividing the
equationx^6 +x^5 +x^4 +x^3 +x^2 +x+ 1 =0 through byx^3 and substitutingyin it, we
obtain the cubic equation
y^3 +y^2 − 2 y− 1 = 0.The numbers 2 cos^27 π, 2 cos^47 π, and 2 cos^87 πare the three roots of this equation. The
simpler task is fulfilled.
But the problem asks us to find the sum of the cube roots of these numbers. Looking
at symmetric polynomials, we have
X^3 +Y^3 +Z^3 − 3 XY Z=(X+Y+Z)^3 − 3 (X+Y+Z)(XY+YZ+ZX)and
X^3 Y^3 +Y^3 Z^3 +Z^3 X^3 − 3 (XY Z)^2 =(XY+YZ+XZ)^3
− 3 XY Z(X+Y+Z)(XY+YZ+ZX).BecauseX^3 ,Y^3 ,Z^3 are the roots of the equationy^3 +y^2 − 2 y− 1 =0, by Viète’s
relations,X^3 Y^3 Z^3 =1, soXY Z =^3
√
1 =1, and alsoX^3 +Y^3 +Z^3 =−1 and
X^3 Y^3 +X^3 Z^3 +Y^3 Z^3 =−2. In the above two equalities we now know the left-hand
sides. The equalities become a system of two equations in the unknownsu=X+Y+Z
andv=XY+YZ+ZX, namely
u^3 − 3 uv=− 4 ,
v^3 − 3 uv=− 5.Writing the two equations asu^3 = 3 uv−4 andv^3 = 3 uv−5 and multiplying them,
we obtain(uv)^3 = 9 (uv)^2 − 27 uv+20. With the substitutionm=uvthis becomes
m^3 − 3 m^3 + 27 m− 20 =0, or(m− 3 )^3 + 7 =0. This equation has the unique solution
m= 3 −^3
√
- Henceu=^3
√
3 m− 4 =^3√
5 − 33
√
- We conclude that
3√
cos
2 π
7+^3
√
cos
4 π
7+^3
√
cos
8 π
7=X+Y+Z=
1
√ 32 u=
3√
1
2
( 5 − 33
√
7 ),
as desired.
All problems below can be solved using Viète’s relations.154.Find the zeros of the polynomial
P(x)=x^4 − 6 x^3 + 18 x^2 − 30 x+ 25
knowing that the sum of two of them is 4.