Summarize Timeline Top Qs Fact Check Los lados del triángulo heptagonal a < b < c coinciden respectivamente con el lado del heptágono regular, diagonal más corta y diagonal más larga. Satisfacen que[ 3] : Lemma 1
a
2
=
c
(
c
−
b
)
,
b
2
=
a
(
c
+
a
)
,
c
2
=
b
(
a
+
b
)
,
1
a
=
1
b
+
1
c
{\displaystyle {\begin{aligned}a^{2}&=c(c-b),\\[5pt]b^{2}&=a(c+a),\\[5pt]c^{2}&=b(a+b),\\[5pt]{\frac {1}{a}}&={\frac {1}{b}}+{\frac {1}{c}}\end{aligned}}}
(la última[ 2] : p. 13 ecuación óptica ) y por lo tanto
a
b
+
a
c
=
b
c
,
{\displaystyle ab+ac=bc,}
y[ 3] : Coro. 2
b
3
+
2
b
2
c
−
b
c
2
−
c
3
=
0
,
{\displaystyle b^{3}+2b^{2}c-bc^{2}-c^{3}=0,}
c
3
−
2
c
2
a
−
c
a
2
+
a
3
=
0
,
{\displaystyle c^{3}-2c^{2}a-ca^{2}+a^{3}=0,}
a
3
−
2
a
2
b
−
a
b
2
+
b
3
=
0.
{\displaystyle a^{3}-2a^{2}b-ab^{2}+b^{3}=0.}
Por lo tanto, -b /c , c /a y a /b satisfacen la ecuación cúbica
t
3
−
2
t
2
−
t
+
1
=
0.
{\displaystyle t^{3}-2t^{2}-t+1=0.}
La relación entre los lados es
b
=
2
cos
(
π
7
)
⋅
a
,
c
=
(
1
+
2
cos
(
2
π
7
)
)
⋅
a
.
{\displaystyle b=2\cos \left({\frac {\pi }{7}}\right)\cdot a,\qquad c=\left(1+2\cos \left({\frac {2\pi }{7}}\right)\right)\cdot a.}
y las raíces de esta ecuación son:
{
t
1
=
1
−
2
cos
(
π
7
)
t
2
=
1
+
2
cos
(
2
π
7
)
t
3
=
4
cos
(
2
π
7
)
cos
(
3
π
7
)
{\displaystyle {\begin{cases}t_{1}=1-2\cos \left({\frac {\pi }{7}}\right)\\t_{2}=1+2\cos \left({\frac {2\pi }{7}}\right)\\t_{3}=4\cos \left({\frac {2\pi }{7}}\right)\cos \left({\frac {3\pi }{7}}\right)\end{cases}}}
[ 4]
a
2
b
c
,
−
b
2
c
a
,
−
c
2
a
b
{\displaystyle {\frac {a^{2}}{bc}},\quad -{\frac {b^{2}}{ca}},\quad -{\frac {c^{2}}{ab}}}
satisface la ecuación cúbica
t
3
+
4
t
2
+
3
t
−
1
=
0
{\displaystyle t^{3}+4t^{2}+3t-1=0}
y las raíces de esta ecuación son:
{
t
1
=
−
1
−
2
cos
(
π
7
)
t
2
=
−
1
+
2
cos
(
2
π
7
)
t
3
=
4
cos
(
2
π
7
)
cos
(
3
π
7
)
−
2
{\displaystyle {\begin{cases}t_{1}=-1-2\cos \left({\frac {\pi }{7}}\right)\\t_{2}=-1+2\cos \left({\frac {2\pi }{7}}\right)\\t_{3}=4\cos \left({\frac {2\pi }{7}}\right)\cos \left({\frac {3\pi }{7}}\right)-2\end{cases}}}
También se tiene que[ 4]
a
3
b
c
2
,
−
b
3
c
a
2
,
c
3
a
b
2
{\displaystyle {\frac {a^{3}}{bc^{2}}},\quad -{\frac {b^{3}}{ca^{2}}},\quad {\frac {c^{3}}{ab^{2}}}}
satisface la ecuación cúbica
t
3
−
t
2
−
9
t
+
1
=
0
{\displaystyle t^{3}-t^{2}-9t+1=0}
y las raíces de esta ecuación son:
{
t
1
=
1
−
4
cos
(
π
7
)
t
2
=
1
+
4
cos
(
2
π
7
)
t
3
=
8
cos
(
2
π
7
)
cos
(
3
π
7
)
−
1
{\displaystyle {\begin{cases}t_{1}=1-4\cos \left({\frac {\pi }{7}}\right)\\t_{2}=1+4\cos \left({\frac {2\pi }{7}}\right)\\t_{3}=8\cos \left({\frac {2\pi }{7}}\right)\cos \left({\frac {3\pi }{7}}\right)-1\end{cases}}}
Así mismo, los valores[ 4]
a
3
b
2
c
,
b
3
c
2
a
,
−
c
3
a
2
b
{\displaystyle {\frac {a^{3}}{b^{2}c}},\quad {\frac {b^{3}}{c^{2}a}},\quad -{\frac {c^{3}}{a^{2}b}}}
satisfacen la ecuación cúbica
t
3
+
5
t
2
−
8
t
+
1
=
0
{\displaystyle t^{3}+5t^{2}-8t+1=0}
y las raíces de esta ecuación son:
{
t
1
=
−
2
[
cos
(
π
7
)
+
2
cos
(
2
π
7
)
+
1
]
t
2
=
6
cos
(
π
7
)
−
2
cos
(
2
π
7
)
−
3
t
3
=
2
[
3
cos
(
2
π
7
)
−
2
cos
(
π
7
)
]
{\displaystyle {\begin{cases}t_{1}=-2\left[\cos \left({\frac {\pi }{7}}\right)+2\cos \left({\frac {2\pi }{7}}\right)+1\right]\\t_{2}=6\cos \left({\frac {\pi }{7}}\right)-2\cos \left({\frac {2\pi }{7}}\right)-3\\t_{3}=2\left[3\cos \left({\frac {2\pi }{7}}\right)-2\cos \left({\frac {\pi }{7}}\right)\right]\end{cases}}}
También se tiene que[ 2] : p. 14
b
2
−
a
2
=
a
c
,
{\displaystyle b^{2}-a^{2}=ac,}
c
2
−
b
2
=
a
b
,
{\displaystyle c^{2}-b^{2}=ab,}
a
2
−
c
2
=
−
b
c
,
{\displaystyle a^{2}-c^{2}=-bc,}
y[ 2] : p. 15
b
2
a
2
+
c
2
b
2
+
a
2
c
2
=
5.
{\displaystyle {\frac {b^{2}}{a^{2}}}+{\frac {c^{2}}{b^{2}}}+{\frac {a^{2}}{c^{2}}}=5.}
Por otro lado[ 4]
a
b
−
b
c
+
c
a
=
0
,
{\displaystyle ab-bc+ca=0,}
a
3
b
−
b
3
c
+
c
3
a
=
0
,
{\displaystyle a^{3}b-b^{3}c+c^{3}a=0,}
a
4
b
+
b
4
c
−
c
4
a
=
0
,
{\displaystyle a^{4}b+b^{4}c-c^{4}a=0,}
a
11
b
3
−
b
11
c
3
+
c
11
a
3
=
0.
{\displaystyle a^{11}b^{3}-b^{11}c^{3}+c^{11}a^{3}=0.}
No hay otro par de números (m, n ), tales que m, n > 0 y que m, n <2000, que cumplan [cita requerida
a
m
b
n
±
b
m
c
n
±
c
m
a
n
=
0.
{\displaystyle a^{m}b^{n}\pm b^{m}c^{n}\pm c^{m}a^{n}=0.}
Las alturas h a h b h c
h
a
=
h
b
+
h
c
{\displaystyle h_{a}=h_{b}+h_{c}}
[ 2] : p. 13 y
h
a
2
+
h
b
2
+
h
c
2
=
a
2
+
b
2
+
c
2
2
.
{\displaystyle h_{a}^{2}+h_{b}^{2}+h_{c}^{2}={\frac {a^{2}+b^{2}+c^{2}}{2}}.}
[ 2] : p. 14 La altura desde el lado b (ángulo opuesto B ) es la mitad de la bisectriz del ángulo interno
w
A
{\displaystyle w_{A}}
A :[ 2] : p. 19
2
h
b
=
w
A
.
{\displaystyle 2h_{b}=w_{A}.}
Aquí el ángulo A es el ángulo más pequeño y B es el segundo ángulo más pequeño.
Se tienen las siguientes propiedades de las bisectrices
w
A
,
w
B
,
{\displaystyle w_{A},w_{B},}
w
C
{\displaystyle w_{C}}
A, B y C respectivamente:[ 2] : p. 16
w
A
=
b
+
c
,
{\displaystyle w_{A}=b+c,}
w
B
=
c
−
a
,
{\displaystyle w_{B}=c-a,}
w
C
=
b
−
a
.
{\displaystyle w_{C}=b-a.}
El área del triángulo es[ 5]
A
=
7
4
R
2
,
{\displaystyle A={\frac {\sqrt {7}}{4}}R^{2},}
donde R es el circunradio del triángulo.
Se tiene que[ 2] : p. 12
a
2
+
b
2
+
c
2
=
7
R
2
.
{\displaystyle a^{2}+b^{2}+c^{2}=7R^{2}.}
También se tiene que[ 6]
a
4
+
b
4
+
c
4
=
21
R
4
.
{\displaystyle a^{4}+b^{4}+c^{4}=21R^{4}.}
a
6
+
b
6
+
c
6
=
70
R
6
.
{\displaystyle a^{6}+b^{6}+c^{6}=70R^{6}.}
La relación
r
R
=
2
cos
(
π
7
)
−
3
2
{\displaystyle {\frac {r}{R}}=2\cos \left({\frac {\pi }{7}}\right)-{\frac {3}{2}}}
inradio respecto al circunradio es la solución positiva de la ecuación cúbica[ 5]
8
x
3
+
28
x
2
+
14
x
−
7
=
0
{\displaystyle 8x^{3}+28x^{2}+14x-7=0}
siendo las otras dos raíces de esta ecuación
2
cos
(
3
π
7
)
−
3
2
{\displaystyle 2\cos \left({\frac {3\pi }{7}}\right)-{\frac {3}{2}}}
2
cos
(
5
π
7
)
−
3
2
{\displaystyle 2\cos \left({\frac {5\pi }{7}}\right)-{\frac {3}{2}}}
La relación
r
a
+
r
b
+
r
c
R
=
2
cos
(
π
7
)
+
5
2
{\displaystyle {\frac {r_{a}+r_{b}+r_{c}}{R}}=2\cos \left({\frac {\pi }{7}}\right)+{\frac {5}{2}}}
8
x
3
−
68
x
2
+
174
x
−
127
=
0
{\displaystyle 8x^{3}-68x^{2}+174x-127=0}
siendo las otras dos raíces de esta ecuación
2
cos
(
3
π
7
)
+
5
2
{\displaystyle 2\cos \left({\frac {3\pi }{7}}\right)+{\frac {5}{2}}}
2
cos
(
5
π
7
)
+
5
2
{\displaystyle 2\cos \left({\frac {5\pi }{7}}\right)+{\frac {5}{2}}}
La relación
1
r
a
+
1
r
b
+
1
r
c
R
=
4
cos
(
2
π
7
)
{\displaystyle {\frac {{\frac {1}{r_{a}}}+{\frac {1}{r_{b}}}+{\frac {1}{r_{c}}}}{R}}=4\cos \left({\frac {2\pi }{7}}\right)}
x
3
+
2
x
2
−
8
x
−
8
=
0
{\displaystyle x^{3}+2x^{2}-8x-8=0}
siendo las otras dos raíces de esta ecuación
4
cos
(
4
π
7
)
{\displaystyle 4\cos \left({\frac {4\pi }{7}}\right)}
4
cos
(
6
π
7
)
{\displaystyle 4\cos \left({\frac {6\pi }{7}}\right)}
Además,[ 2] : p. 15
1
a
2
+
1
b
2
+
1
c
2
=
2
R
2
.
{\displaystyle {\frac {1}{a^{2}}}+{\frac {1}{b^{2}}}+{\frac {1}{c^{2}}}={\frac {2}{R^{2}}}.}
También se tiene que[ 6]
1
a
4
+
1
b
4
+
1
c
4
=
2
R
4
.
{\displaystyle {\frac {1}{a^{4}}}+{\frac {1}{b^{4}}}+{\frac {1}{c^{4}}}={\frac {2}{R^{4}}}.}
1
a
6
+
1
b
6
+
1
c
6
=
17
7
R
6
.
{\displaystyle {\frac {1}{a^{6}}}+{\frac {1}{b^{6}}}+{\frac {1}{c^{6}}}={\frac {17}{7R^{6}}}.}
En general para todos los enteros n ,
a
2
n
+
b
2
n
+
c
2
n
=
g
(
n
)
(
2
R
)
2
n
{\displaystyle a^{2n}+b^{2n}+c^{2n}=g(n)(2R)^{2n}}
donde
g
(
−
1
)
=
8
,
g
(
0
)
=
3
,
g
(
1
)
=
7
{\displaystyle g(-1)=8,\quad g(0)=3,\quad g(1)=7}
y
g
(
n
)
=
7
g
(
n
−
1
)
−
14
g
(
n
−
2
)
+
7
g
(
n
−
3
)
.
{\displaystyle g(n)=7g(n-1)-14g(n-2)+7g(n-3).}
Así mismo[ 6]
2
b
2
−
a
2
=
7
b
R
,
2
c
2
−
b
2
=
7
c
R
,
2
a
2
−
c
2
=
−
7
a
R
.
{\displaystyle 2b^{2}-a^{2}={\sqrt {7}}bR,\quad 2c^{2}-b^{2}={\sqrt {7}}cR,\quad 2a^{2}-c^{2}=-{\sqrt {7}}aR.}
También se tiene que[ 4]
a
3
c
+
b
3
a
−
c
3
b
=
−
7
R
4
,
{\displaystyle a^{3}c+b^{3}a-c^{3}b=-7R^{4},}
a
4
c
−
b
4
a
+
c
4
b
=
7
7
R
5
,
{\displaystyle a^{4}c-b^{4}a+c^{4}b=7{\sqrt {7}}R^{5},}
a
11
c
3
+
b
11
a
3
−
c
11
b
3
=
−
7
3
17
R
14
.
{\displaystyle a^{11}c^{3}+b^{11}a^{3}-c^{11}b^{3}=-7^{3}17R^{14}.}
El exradio r a a es igual al radio de la circunferencia de los nueve puntos del triángulo heptagonal.[ 2] : p. 15
![{\displaystyle {\begin{aligned}a^{2}&=c(c-b),\\[5pt]b^{2}&=a(c+a),\\[5pt]c^{2}&=b(a+b),\\[5pt]{\frac {1}{a}}&={\frac {1}{b}}+{\frac {1}{c}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63107bd555aab31d4dbd45b3bea63fcb044aeb43)
![{\displaystyle {\begin{cases}t_{1}=-2\left[\cos \left({\frac {\pi }{7}}\right)+2\cos \left({\frac {2\pi }{7}}\right)+1\right]\\t_{2}=6\cos \left({\frac {\pi }{7}}\right)-2\cos \left({\frac {2\pi }{7}}\right)-3\\t_{3}=2\left[3\cos \left({\frac {2\pi }{7}}\right)-2\cos \left({\frac {\pi }{7}}\right)\right]\end{cases}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4adbd5db36fdb12f6fcfbde16edc40037bf88b6)
![{\displaystyle {\sqrt[{3}]{2\operatorname {sen}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{2\operatorname {sen}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{2\operatorname {sen}({\frac {8\pi }{7}})}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/944fa265cc4a9e01179032bc93e04f3bad74375c)
![{\displaystyle {\text{.......}}\left(-{\sqrt[{18}]{7}}\right){\sqrt[{3}]{-{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c8a93053ce8ad0eb9ddcb3cc47d03aeb44597eb1)
![{\displaystyle {\frac {1}{\sqrt[{3}]{2\operatorname {sen}({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{2\operatorname {sen}({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{2\operatorname {sen}({\frac {8\pi }{7}})}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f48096e10b7da3bcd8754842b04dd84755e91613)
![{\displaystyle {\text{.......}}\left(-{\frac {1}{\sqrt[{18}]{7}}}\right){\sqrt[{3}]{6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee8a1e5f7a4a7646390f002f725c8937e8282894)
![{\displaystyle {\sqrt[{3}]{4\operatorname {sen} ^{2}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{4\operatorname {sen} ^{2}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{4\operatorname {sen} ^{2}({\frac {8\pi }{7}})}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/64d8d1922a51de41472b406b267ec8d1edb9e0f4)
![{\displaystyle {\text{.......}}\left({\sqrt[{18}]{49}}\right){\sqrt[{3}]{{\sqrt[{3}]{49}}+6+3\left({\sqrt[{3}]{12+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{11+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aba2a19e87ed078ce46a84894486a90ae81b9a1f)
![{\displaystyle {\frac {1}{\sqrt[{3}]{4\operatorname {sen} ^{2}({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{4\operatorname {sen} ^{2}({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{4\operatorname {sen} ^{2}({\frac {8\pi }{7}})}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ad5bf445b03e9e6b31f71598805dccbf20a7e195)
![{\displaystyle {\text{.......}}\left({\frac {1}{\sqrt[{18}]{49}}}\right){\sqrt[{3}]{2{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{12+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{11+3({\sqrt[{3}]{49}}+2{\sqrt[{3}]{7}})}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba0013568e9972e6aae1aec3e8c85493e626fdcd)
![{\displaystyle {\sqrt[{3}]{2\cos({\frac {2\pi }{7}})}}+{\sqrt[{3}]{2\cos({\frac {4\pi }{7}})}}+{\sqrt[{3}]{2\cos({\frac {8\pi }{7}})}}={\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b314c223d0be63a0f72ff996b5fd8fcc12d7e7f0)
![{\displaystyle {\frac {1}{\sqrt[{3}]{2\cos({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{2\cos({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{2\cos({\frac {8\pi }{7}})}}}={\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b166c9b5596c43166f06dae47417e000f8f47807)
![{\displaystyle {\sqrt[{3}]{4\cos ^{2}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{4\cos ^{2}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{4\cos ^{2}({\frac {8\pi }{7}})}}={\sqrt[{3}]{11+3(2{\sqrt[{3}]{7}}+{\sqrt[{3}]{49}})}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/195af596d79a11379a62885b6934e92f9e028386)
![{\displaystyle {\frac {1}{\sqrt[{3}]{4\cos ^{2}({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{4\cos ^{2}({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{4\cos ^{2}({\frac {8\pi }{7}})}}}={\sqrt[{3}]{12+3(2{\sqrt[{3}]{7}}+{\sqrt[{3}]{49}})}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9b69e7f053c8c3e7301b9043df9d030c894ffe8b)
![{\displaystyle {\sqrt[{3}]{\tan({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\tan({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\tan({\frac {8\pi }{7}})}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a52f81bb6d9f9c7d2edf6a15127b6509d173c2a)
![{\displaystyle {\text{.......}}\left(-{\sqrt[{18}]{7}}\right){\sqrt[{3}]{{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}+{\sqrt[{3}]{-3+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d40e8336542726b021dbc22dfc8df6f1f0360223)
![{\displaystyle {\frac {1}{\sqrt[{3}]{\tan({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{\tan({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{\tan({\frac {8\pi }{7}})}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93fae279c3c18761c02980f1de1f56add9b85ac2)
![{\displaystyle {\text{.......}}\left(-{\frac {1}{\sqrt[{18}]{7}}}\right){\sqrt[{3}]{-{\sqrt[{3}]{49}}+6+3\left({\sqrt[{3}]{5+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}+{\sqrt[{3}]{-3+3({\sqrt[{3}]{7}}-{\sqrt[{3}]{49}})}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc97fe95f2898ee704610d68aa4b316420511852)
![{\displaystyle {\sqrt[{3}]{\tan ^{2}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\tan ^{2}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\tan ^{2}({\frac {8\pi }{7}})}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5fb410a07939f6d720b679c008c5e43729fc9998)
![{\displaystyle {\text{.......}}\left({\sqrt[{18}]{49}}\right){\sqrt[{3}]{3{\sqrt[{3}]{49}}+6+3\left({\sqrt[{3}]{89+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{25+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c44c9c7171b3fe668d604b31cdd4770b74662fd)
![{\displaystyle {\frac {1}{\sqrt[{3}]{\tan ^{2}({\frac {2\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{\tan ^{2}({\frac {4\pi }{7}})}}}+{\frac {1}{\sqrt[{3}]{\tan ^{2}({\frac {8\pi }{7}})}}}=}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ae013c02c7bfa040e9e92343e56a699900fcce0)
![{\displaystyle {\text{.......}}\left({\frac {1}{\sqrt[{18}]{49}}}\right){\sqrt[{3}]{5{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{89+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}+{\sqrt[{3}]{25+3(3{\sqrt[{3}]{49}}+5{\sqrt[{3}]{7}})}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7f12b27533fca50aee78833799b2de718f13068)
![{\displaystyle {\sqrt[{3}]{\cos({\frac {2\pi }{7}})/\cos({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos({\frac {4\pi }{7}})/\cos({\frac {8\pi }{7}})}}+{\sqrt[{3}]{\cos({\frac {8\pi }{7}})/\cos({\frac {2\pi }{7}})}}=-{\sqrt[{3}]{7}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e5c104a4280b7a96ff7a0675b28b99ab5ae46262)
![{\displaystyle {\sqrt[{3}]{\cos({\frac {4\pi }{7}})/\cos({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\cos({\frac {8\pi }{7}})/\cos({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos({\frac {2\pi }{7}})/\cos({\frac {8\pi }{7}})}}=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/63365b601c7fdc95ac2ecf405bf3f8fe5a016364)
![{\displaystyle {\sqrt[{3}]{2\operatorname {sen}({2\pi }{7}}}+{\sqrt[{3}]{2\operatorname {sen}({4\pi }{7}}}+{\sqrt[{3}]{2\operatorname {sen}({8\pi }{7}}}=\left(-{\sqrt[{18}]{7}}\right){\sqrt[{3}]{-{\sqrt[{3}]{7}}+6+3\left({\sqrt[{3}]{5-3{\sqrt[{3}]{7}}}}+{\sqrt[{3}]{4-3{\sqrt[{3}]{7}}}}\right)}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/307026a941c7352dfed95f0c14aaa465015817fd)
![{\displaystyle {\sqrt[{3}]{\cos ^{4}({\frac {4\pi }{7}})/\cos({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\cos ^{4}({\frac {8\pi }{7}})/\cos({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos ^{4}({\frac {2\pi }{7}})/\cos({\frac {8\pi }{7}})}}=-{\sqrt[{3}]{49}}/2.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc578e6203860ba7806ed05aab4c5ae03fc9c07f)
![{\displaystyle {\sqrt[{3}]{\cos ^{5}({\frac {2\pi }{7}})/\cos ^{2}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos ^{5}({\frac {4\pi }{7}})/\cos ^{2}({\frac {8\pi }{7}})}}+{\sqrt[{3}]{\cos ^{5}({\frac {8\pi }{7}})/\cos ^{2}({\frac {2\pi }{7}})}}=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/932f7f96c30f5d525cfc095ad31c67c2ab91b322)
![{\displaystyle {\sqrt[{3}]{\cos ^{5}({\frac {4\pi }{7}})/\cos ^{2}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\cos ^{5}({\frac {8\pi }{7}})/\cos ^{2}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos ^{5}({\frac {2\pi }{7}})/\cos ^{2}({\frac {9\pi }{7}})}}=-3*{\sqrt[{3}]{7}}/2.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6bd72754057556b76216a587321ce081449f8094)
![{\displaystyle {\sqrt[{3}]{\cos ^{14}({\frac {2\pi }{7}})/\cos ^{5}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos ^{14}({\frac {4\pi }{7}})/\cos ^{5}({\frac {8\pi }{7}})}}+{\sqrt[{3}]{\cos ^{14}({\frac {8\pi }{7}})/\cos ^{5}({\frac {2\pi }{7}}}}=0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b714fc2f8791a0274bb594df329dffafe897c617)
![{\displaystyle {\sqrt[{3}]{\cos ^{14}({\frac {4\pi }{7}})/\cos ^{5}({\frac {2\pi }{7}})}}+{\sqrt[{3}]{\cos ^{14}({\frac {8\pi }{7}})/\cos ^{5}({\frac {4\pi }{7}})}}+{\sqrt[{3}]{\cos ^{14}({\frac {2\pi }{7}})/\cos ^{5}({\frac {8\pi }{7}})}}=-61*{\sqrt[{3}]{7}}/8.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0e9412d68dbce7ac0c38e9580d731a5281295f43)
