육차 방정식 (Sextic equation )은 최고차항의 차수가 6인 다항 방정식 을 뜻한다. 일반적인 형태는
a x 6 + b x 5 + c x 4 + d x 3 + e x 2 + f x + g = 0 , a ≠ 0 {\displaystyle ax^{6}+bx^{5}+cx^{4}+dx^{3}+ex^{2}+fx+g=0,\,a\neq 0} 와 같다. 여기에서 a , b , c , d , e , f {\displaystyle a,b,c,d,e,f} x 6 , x 5 , x 4 , x 3 , x 2 , x {\displaystyle x^{6},x^{5},x^{4},x^{3},x^{2},x} 계수 라고 한다. g {\displaystyle g}
또한 차수가 홀수인 경우를 기수차라고 하고 짝수인 경우를 우수차라고 할 때, 육차방정식은 우수차 방정식이다.
대수학의 기본 정리 (fundamental theorem of algebra)에 의해 복소수 범위에서 육차방정식의 해는 항상 존재한다. 다만, 아벨 과 갈루아 는 기하적인 면을 배제하고 계수만 가지고 5차 이상의 방정식에서는 표현할 수 없다는 것을 증명했다. 즉, 그 근(해)을 임의의 계수들을 가지고서는 유한번의 사칙연산과 제곱근 연산으로 표현할 수 없다는 것이다. 여기에 대한 아벨 과 루피니 의 증명은 아벨-루피니 정리 이다.
[ 편집 ] 복(複) 삼차방정식
육차 방정식 중 짝수 차수만 있는 방정식을 복삼차방정식 (Bicubic equations)이라고 한다.
x 2 = X {\displaystyle x^{2}=X} 치환 해 삼차방정식의 풀이를 이용해 푼다. a x 6 + b x 4 + c x 2 + d = 0 , X = x 2 {\displaystyle ax^{6}+bx^{4}+cx^{2}+d=0\;,\;X=x^{2}} a X 3 + b X 2 + c X + d = 0 {\displaystyle aX^{3}+bX^{2}+cX+d=0\;} 이 방법을 이용하여 a x 6 + b x 3 + c {\displaystyle ax^{6}+bx^{3}+c} x 3 {\displaystyle x^{3}}
상반방정식
x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 = 0 {\displaystyle x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1=0\,} x n 2 {\displaystyle x^{n \over 2}\;} x 6 x 3 + x 5 x 3 + x 4 x 3 + x 3 x 3 + x 2 x 3 + x x 3 + 1 x 3 = 0 {\displaystyle {x^{6} \over x^{3}}+{x^{5} \over x^{3}}+{x^{4} \over x^{3}}+{x^{3} \over x^{3}}+{x^{2} \over x^{3}}+{x \over x^{3}}+{1 \over x^{3}}=0\,} x 3 + x 2 + x + 1 + 1 x + 1 x 2 + 1 x 3 = 0 {\displaystyle x^{3}+x^{2}+x+1+{1 \over x}+{1 \over x^{2}}+{1 \over x^{3}}=0\,} ( x 3 + 1 x 3 ) + ( x 2 + 1 x 2 ) + ( x + 1 x ) + 1 = 0 {\displaystyle \left(x^{3}+{1 \over x^{3}}\right)+\left(x^{2}+{1 \over x^{2}}\right)+\left(x+{1 \over x}\right)+1=0} x + 1 x = z {\displaystyle x+{1 \over x}=z\qquad } z 3 + z 2 − 2 z − 1 = 0 {\displaystyle z^{3}+z^{2}-2z-1=0\qquad } z = x + 1 x {\displaystyle z=x+{1 \over x}\qquad } x 2 + 1 x = z {\displaystyle {{x^{2}+1} \over x}=z} x 2 + 1 = z x {\displaystyle {x^{2}+1}=zx} x 2 − z x + 1 = 0 {\displaystyle {x^{2}}-zx+1=0} 이것으로 이차방정식을 풀면, 각각 2개씩의 근, 즉 총 6개의 근을 구하게 된다.
이항방정식
x 6 ± a = 0 {\displaystyle x^{6}\pm a=0} a {\displaystyle a} ω {\displaystyle \omega } x 6 = x 2 + 2 x + 1 {\displaystyle x^{6}=x^{2}+2x+1\;} x 6 = x 4 + x + 1 {\displaystyle \!\ x^{6}=x^{4}+x+1} 다음의 육차 방정식들은
1 2 + 1 6 23 3 3 + 1 2 − 1 6 23 3 3 = 1.324717957244746025960908854... {\displaystyle {\sqrt[{3}]{{{1} \over {2}}+{{1} \over {6}}{\sqrt {{23} \over {3}}}}}+{\sqrt[{3}]{{{1} \over {2}}-{{1} \over {6}}{\sqrt {{23} \over {3}}}}}=1.324717957244746025960908854...} 플라스틱 수 가 해이다.
육차 방정식의 판별식 은 246개항으로 이루어져 있다.
소행렬식 의 라플라스 전개 로 실베스터 행렬 의 종결식을 사용한 육차방정식의 판별식 유도가 가능하다.
6차방정식 a x 6 + b x 5 + c x 4 + d x 3 + e x 2 + f x + g = 0 {\displaystyle \textstyle ax^{6}+bx^{5}+cx^{4}+dx^{3}+ex^{2}+fx+g=0} α , β , γ , δ , ϵ , ζ {\displaystyle \textstyle \alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta }
( x − α ) ( x − β ) ( x − γ ) ( x − δ ) ( x − ϵ ) ( x − ζ ) = 0 {\displaystyle (x-\alpha )(x-\beta )(x-\gamma )(x-\delta )(x-\epsilon )(x-\zeta )=0} x 6 − ( α + β + γ + δ + ϵ + ζ ) x 5 + ( α β + α γ + α δ + α ϵ + α ζ + β γ + β δ + β ϵ + β ζ + γ δ + γ ϵ + γ ζ + δ ϵ + δ ζ + ϵ ζ ) x 4 {\displaystyle x^{6}-(\alpha +\beta +\gamma +\delta +\epsilon +\zeta )x^{5}+(\alpha \beta +\alpha \gamma +\alpha \delta +\alpha \epsilon +\alpha \zeta +\beta \gamma +\beta \delta +\beta \epsilon +\beta \zeta +\gamma \delta +\gamma \epsilon +\gamma \zeta +\delta \epsilon +\delta \zeta +\epsilon \zeta )x^{4}} − ( α β γ + α β δ + α β ϵ + α β ζ + α γ δ + α γ ϵ + α γ ζ + α δ ϵ + α δ ζ + α ϵ ζ + β γ δ + β γ ϵ + β γ ζ + β ϵ ζ + β δ ϵ + β δ ζ + γ δ ϵ + γ δ ζ + γ ϵ ζ + δ ϵ ζ ) x 3 {\displaystyle -(\alpha \beta \gamma +\alpha \beta \delta +\alpha \beta \epsilon +\alpha \beta \zeta +\alpha \gamma \delta +\alpha \gamma \epsilon +\alpha \gamma \zeta +\alpha \delta \epsilon +\alpha \delta \zeta +\alpha \epsilon \zeta +\beta \gamma \delta +\beta \gamma \epsilon +\beta \gamma \zeta +\beta \epsilon \zeta +\beta \delta \epsilon +\beta \delta \zeta +\gamma \delta \epsilon +\gamma \delta \zeta +\gamma \epsilon \zeta +\delta \epsilon \zeta )x^{3}}
+ ( α β γ δ + α β γ ϵ + α β γ ζ + α β δ ϵ + α β δ ζ + α β ϵ ζ + α γ δ ϵ + α γ δ ζ + α γ ϵ ζ + α δ ϵ ζ + β γ δ ϵ + β γ δ ζ + β γ ϵ ζ + β δ ϵ ζ + γ δ ϵ ζ ) x 2 {\displaystyle +(\alpha \beta \gamma \delta +\alpha \beta \gamma \epsilon +\alpha \beta \gamma \zeta +\alpha \beta \delta \epsilon +\alpha \beta \delta \zeta +\alpha \beta \epsilon \zeta +\alpha \gamma \delta \epsilon +\alpha \gamma \delta \zeta +\alpha \gamma \epsilon \zeta +\alpha \delta \epsilon \zeta +\beta \gamma \delta \epsilon +\beta \gamma \delta \zeta +\beta \gamma \epsilon \zeta +\beta \delta \epsilon \zeta +\gamma \delta \epsilon \zeta )x^{2}} − ( α β γ δ ϵ + α β γ δ ζ + α β γ ϵ ζ + α β δ ϵ ζ + α γ δ ϵ ζ + β γ δ ϵ ζ ) x + α β γ δ ϵ ζ = 0 {\displaystyle -(\alpha \beta \gamma \delta \epsilon +\alpha \beta \gamma \delta \zeta +\alpha \beta \gamma \epsilon \zeta +\alpha \beta \delta \epsilon \zeta +\alpha \gamma \delta \epsilon \zeta +\beta \gamma \delta \epsilon \zeta )x+\alpha \beta \gamma \delta \epsilon \zeta =0} 이어서,
α + β + γ + δ + ϵ + ζ = − b a {\displaystyle \alpha +\beta +\gamma +\delta +\epsilon +\zeta =-{b \over a}} α β + α γ + α δ + α ϵ + α ζ + β γ + β δ + β ϵ + β ζ + γ δ + γ ϵ + γ ζ + δ ϵ + δ ζ + ϵ ζ = c a {\displaystyle \alpha \beta +\alpha \gamma +\alpha \delta +\alpha \epsilon +\alpha \zeta +\beta \gamma +\beta \delta +\beta \epsilon +\beta \zeta +\gamma \delta +\gamma \epsilon +\gamma \zeta +\delta \epsilon +\delta \zeta +\epsilon \zeta ={c \over a}} α β γ + α β δ + α β ϵ + α β ζ + α γ δ + α γ ϵ + α γ ζ + α δ ϵ + α δ ζ + α ϵ ζ + β γ δ + β γ ϵ + β γ ζ + β ϵ ζ + β δ ϵ + β δ ζ + γ δ ϵ + γ δ ζ + γ ϵ ζ + δ ϵ ζ = − d a {\displaystyle \alpha \beta \gamma +\alpha \beta \delta +\alpha \beta \epsilon +\alpha \beta \zeta +\alpha \gamma \delta +\alpha \gamma \epsilon +\alpha \gamma \zeta +\alpha \delta \epsilon +\alpha \delta \zeta +\alpha \epsilon \zeta +\beta \gamma \delta +\beta \gamma \epsilon +\beta \gamma \zeta +\beta \epsilon \zeta +\beta \delta \epsilon +\beta \delta \zeta +\gamma \delta \epsilon +\gamma \delta \zeta +\gamma \epsilon \zeta +\delta \epsilon \zeta =-{d \over a}} α β γ δ + α β γ ϵ + α β γ ζ + α β δ ϵ + α β δ ζ + α β ϵ ζ + α γ δ ϵ + α γ δ ζ + α γ ϵ ζ + α δ ϵ ζ + β γ δ ϵ + β γ δ ζ + β γ ϵ ζ + β δ ϵ ζ + γ δ ϵ ζ = e a {\displaystyle \alpha \beta \gamma \delta +\alpha \beta \gamma \epsilon +\alpha \beta \gamma \zeta +\alpha \beta \delta \epsilon +\alpha \beta \delta \zeta +\alpha \beta \epsilon \zeta +\alpha \gamma \delta \epsilon +\alpha \gamma \delta \zeta +\alpha \gamma \epsilon \zeta +\alpha \delta \epsilon \zeta +\beta \gamma \delta \epsilon +\beta \gamma \delta \zeta +\beta \gamma \epsilon \zeta +\beta \delta \epsilon \zeta +\gamma \delta \epsilon \zeta ={e \over a}} α β γ δ ϵ + α β γ δ ζ + α β γ ϵ ζ + α β δ ϵ ζ + α γ δ ϵ ζ + β γ δ ϵ ζ = − f a {\displaystyle \alpha \beta \gamma \delta \epsilon +\alpha \beta \gamma \delta \zeta +\alpha \beta \gamma \epsilon \zeta +\alpha \beta \delta \epsilon \zeta +\alpha \gamma \delta \epsilon \zeta +\beta \gamma \delta \epsilon \zeta =-{f \over a}} α β γ δ ϵ ζ = g a {\displaystyle \alpha \beta \gamma \delta \epsilon \zeta ={g \over a}} 특히 각 항( x 6 , x 5 , x 4 , x 3 , x 2 , x , g {\displaystyle x^{6},x^{5},x^{4},x^{3},x^{2},x,g} 조합 의 경우의 수 로 따져 볼 수 있다.
6차방정식에 존재하는 6개의 근을 예약하여, α , β , γ , δ , ϵ , ζ {\displaystyle \textstyle \alpha ,\beta ,\gamma ,\delta ,\epsilon ,\zeta }
n ! k ! ⋅ ( n − k ) ! = 6 ! 1 ! ⋅ ( 6 − 1 ) ! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 1 ! ⋅ ( 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 ) = 6 1 = 6 {\displaystyle {\frac {n!}{k!\cdot (n-k)!}}={\frac {6!}{1!\cdot (6-1)!}}={{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1} \over {1!\cdot (5\cdot 4\cdot 3\cdot 2\cdot 1)}}={6 \over 1}=6} 2개씩의 조합의 경우의 수는 ,
n ! k ! ⋅ ( n − k ) ! = 6 ! 2 ! ⋅ ( 6 − 2 ) ! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 2 ! ⋅ ( 4 ⋅ 3 ⋅ 2 ⋅ 1 ) = 6 ⋅ 5 2 ⋅ 1 = 30 2 = 15 {\displaystyle {\frac {n!}{k!\cdot (n-k)!}}={\frac {6!}{2!\cdot (6-2)!}}={{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1} \over {2!\cdot (4\cdot 3\cdot 2\cdot 1)}}={{6\cdot 5} \over {2\cdot 1}}={30 \over 2}=15} 3개씩의 조합의 경우의 수는 ,
n ! k ! ⋅ ( n − k ) ! = 6 ! 3 ! ⋅ ( 6 − 3 ) ! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 3 ! ⋅ ( 3 ⋅ 2 ⋅ 1 ) = 6 ⋅ 5 ⋅ 4 3 ⋅ 2 ⋅ 1 = 120 6 = 20 {\displaystyle {\frac {n!}{k!\cdot (n-k)!}}={\frac {6!}{3!\cdot (6-3)!}}={{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1} \over {3!\cdot (3\cdot 2\cdot 1)}}={{6\cdot 5\cdot 4} \over {3\cdot 2\cdot 1}}={120 \over 6}=20} 4개씩의 조합의 경우의 수는 ,
n ! k ! ⋅ ( n − k ) ! = 6 ! 4 ! ⋅ ( 6 − 4 ) ! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 4 ! ⋅ ( 2 ⋅ 1 ) = 6 ⋅ 5 2 ⋅ 1 = 30 2 = 15 {\displaystyle {\frac {n!}{k!\cdot (n-k)!}}={\frac {6!}{4!\cdot (6-4)!}}={{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1} \over {4!\cdot (2\cdot 1)}}={{6\cdot 5} \over {2\cdot 1}}={30 \over 2}=15} 5개씩의 조합의 경우의 수는 ,
n ! k ! ⋅ ( n − k ) ! = 6 ! 5 ! ⋅ ( 6 − 5 ) ! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 5 ! ⋅ ( 1 ) = 6 1 = 6 {\displaystyle {\frac {n!}{k!\cdot (n-k)!}}={\frac {6!}{5!\cdot (6-5)!}}={{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1} \over {5!\cdot (1)}}={{6} \over {1}}={6}} 6개씩의 조합의 경우의 수는 ,
n ! k ! ⋅ ( n − k ) ! = 6 ! 6 ! ⋅ ( 6 − 6 ) ! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 6 ! ⋅ 0 ! = 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 720 720 = 1 {\displaystyle {\frac {n!}{k!\cdot (n-k)!}}={\frac {6!}{6!\cdot (6-6)!}}={{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1} \over {6!\cdot 0!}}={{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1} \over {6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}}={720 \over 720}=1} a x 6 + b x 5 + c x 4 + d x 3 + e x 2 + f x + g = 0 {\displaystyle ax^{6}+bx^{5}+cx^{4}+dx^{3}+ex^{2}+fx+g=0} 다항 방정식에서 양변의 각항들을 해당 방정식의 최고차항( n {\displaystyle n} x {\displaystyle x} a {\displaystyle a} x = y − b n a {\displaystyle \textstyle x=y-{b \over \mathbf {n} a}}
위의 육차 방정식은 다음의 꼴로 정리되겠다. x 6 + b a x 5 + c a x 4 + d a x 3 + e a x 2 + f a x + g a = 0 , x = y − b 6 a {\displaystyle x^{6}+{b \over a}x^{5}+{c \over a}x^{4}+{d \over a}x^{3}+{e \over a}x^{2}+{f \over a}x+{g \over a}=0,\qquad x=y-{b \over \mathbf {6} a}}
y 6 + p y 4 + q y 3 + r y 2 + s y + t = 0 {\displaystyle y^{6}+py^{4}+qy^{3}+ry^{2}+sy+t=0} 여기서 이렇게 압축해서 정리하면 p , q , r , s , t {\displaystyle p,q,r,s,t}
p = ( − 5 b 2 + 12 a c 12 a 2 ) {\displaystyle p=\left({{-5b^{2}+12ac} \over {12a^{2}}}\right)} q = ( 5 b 3 − 18 a b c + 27 a 2 d 27 a 3 ) {\displaystyle q=\left({{5b^{3}-18abc+27a^{2}d} \over {27a^{3}}}\right)} r = ( − 11 b 4 + 72 a c b 2 − 216 a 2 d b + 432 a e 432 a 4 ) {\displaystyle r=\left({{-11b^{4}+72acb^{2}-216a^{2}db+432ae} \over {432a^{4}}}\right)} s = ( − c b 3 + 9 a d b 2 − 36 a 2 e b + 54 a f 54 a 4 ) {\displaystyle s=\left({{-cb^{3}+9adb^{2}-36a^{2}eb+54af} \over {54a^{4}}}\right)} t = ( 14 b 6 + 1 a c b 4 + 6 a 2 d b 3 + 36 a 5 e b 2 − 7776 a 6 f b + 1296 a 5 g 1296 a 6 ) {\displaystyle t=\left({{14b^{6}+1acb^{4}+6a^{2}db^{3}+36a^{5}eb^{2}-7776a^{6}fb+1296a^{5}g} \over {1296a^{6}}}\right)} 여기서 이처럼 오차항이 사라지는 것은 없어진것이 아니라 나머지 다른 항들에게로 자신의 정보들이 나뉘어서 저장된 것이므로 다른 항들을 압축정리하려고 하면 오차항은 다시 나타나게 된다.
이것은 마치 육차항의 계수가 없어진것이 아니라 나머지 항들에게로 자신의 정보를 나누어주고 사라진 것과 같다.
따라서 취른하우스 변형(Tschirnhaus transformation) 정리(차 고차항 압축 정리)는 차고차항 뿐만 아니라 다른 임의의 항들을 1 {\displaystyle 1}
a x 6 + b x 5 + c x 4 + d x 3 + e x 2 + f x + g = 0 {\displaystyle ax^{6}+bx^{5}+cx^{4}+dx^{3}+ex^{2}+fx+g=0} x 6 + b a x 5 + c a x 4 + d a x 3 + e a x 2 + f a x + g a = 0 , {\displaystyle x^{6}+{b \over a}x^{5}+{c \over a}x^{4}+{d \over a}x^{3}+{e \over a}x^{2}+{f \over a}x+{g \over a}=0,} x = y − b 6 a {\displaystyle \qquad x=y-{b \over \mathbf {6} a}} y 6 + p y 4 + q y 3 + r y 2 + s y + t = 0 {\displaystyle y^{6}+py^{4}+qy^{3}+ry^{2}+sy+t=0} p = ( − 5 b 2 + 12 a c 12 a 2 ) {\displaystyle p=\left({{-5b^{2}+12ac} \over {12a^{2}}}\right)} q = ( 5 b 3 − 18 a b c + 27 a 2 d 27 a 3 ) {\displaystyle q=\left({{5b^{3}-18abc+27a^{2}d} \over {27a^{3}}}\right)} r = ( − 11 b 4 + 72 a c b 2 − 216 a 2 d b + 432 a e 432 a 4 ) {\displaystyle r=\left({{-11b^{4}+72acb^{2}-216a^{2}db+432ae} \over {432a^{4}}}\right)} s = ( − c b 3 + 9 a d b 2 − 36 a 2 e b + 54 a f 54 a 4 ) {\displaystyle s=\left({{-cb^{3}+9adb^{2}-36a^{2}eb+54af} \over {54a^{4}}}\right)} t = ( 14 b 6 + 1 a c b 4 + 6 a 2 d b 3 + 36 a 5 e b 2 − 7776 a 6 f b + 1296 a 5 g 1296 a 6 ) {\displaystyle t=\left({{14b^{6}+1acb^{4}+6a^{2}db^{3}+36a^{5}eb^{2}-7776a^{6}fb+1296a^{5}g} \over {1296a^{6}}}\right)} 이어서,
( r y + s 2 r ) 2 = ( r y + s 2 r ) ( r y + s 2 r ) = r y 2 + r s y 2 r + r s y 2 r + ( s 2 r ) 2 = r y 2 + 2 r s y 2 r + ( s 2 r ) 2 {\displaystyle \left({\sqrt {r}}y+{s \over 2{\sqrt {r}}}\right)^{2}=\left({\sqrt {r}}y+{s \over 2{\sqrt {r}}}\right)\left({\sqrt {r}}y+{s \over 2{\sqrt {r}}}\right)=ry^{2}+{{\sqrt {r}}sy \over 2{\sqrt {r}}}+{{\sqrt {r}}sy \over 2{\sqrt {r}}}+\left({s \over 2{\sqrt {r}}}\right)^{2}=ry^{2}+2{{\sqrt {r}}sy \over 2{\sqrt {r}}}+\left({s \over 2{\sqrt {r}}}\right)^{2}} = r y 2 + 2 r s y 2 r + ( s 2 r ) 2 {\displaystyle \;\;\;=ry^{2}+{\cancel {2}}{{\cancel {\sqrt {r}}}sy \over {\cancel {2{\sqrt {r}}}}}+\left({s \over 2{\sqrt {r}}}\right)^{2}} = r y 2 + s y + ( s 2 r ) 2 {\displaystyle \;\;\;=ry^{2}+{sy}+\left({s \over 2{\sqrt {r}}}\right)^{2}} {\displaystyle } ( y 3 + q 2 ) 2 = y 6 + q y 3 2 + q y 3 2 + ( q 2 ) 2 = y 6 + 2 q y 3 2 + ( q 2 ) 2 = y 6 + 2 q y 3 2 + ( q 2 ) 2 {\displaystyle \left(y^{3}+{q \over 2}\right)^{2}=y^{6}+{{qy^{3}} \over 2}+{{qy^{3}} \over 2}+\left({q \over 2}\right)^{2}=y^{6}+2{{qy^{3}} \over 2}+\left({q \over 2}\right)^{2}=y^{6}+{\cancel {2}}{{qy^{3}} \over {\cancel {2}}}+\left({q \over 2}\right)^{2}} = y 6 + q y 3 + ( q 2 ) 2 {\displaystyle =y^{6}+{qy^{3}}+\left({q \over 2}\right)^{2}} ∴ ( y 3 + q 2 ) 2 + p y 4 + ( r y + s 2 r ) 2 + t − ( q 2 ) 2 − ( s 2 r ) 2 = 0 {\displaystyle \therefore \;\;\left(y^{3}+{q \over 2}\right)^{2}+py^{4}+\left({\sqrt {r}}y+{s \over 2{\sqrt {r}}}\right)^{2}+t-\left({q \over 2}\right)^{2}-\left({s \over 2{\sqrt {r}}}\right)^{2}=0} ( y 3 + q 2 ) 2 + ( p y 2 ) 2 + ( r y + s 2 r ) 2 + ( t ) 2 − ( q 2 ) 2 − ( s 2 r ) 2 = 0 {\displaystyle \left(y^{3}+{q \over 2}\right)^{2}+({\sqrt {p}}y^{2})^{2}+\left({\sqrt {r}}y+{s \over 2{\sqrt {r}}}\right)^{2}+({\sqrt {t}})^{2}-\left({q \over 2}\right)^{2}-\left({s \over 2{\sqrt {r}}}\right)^{2}=0} ( y 3 + q 2 ) 2 = A 2 , ( p y 2 ) 2 = B 2 , ( r y + s 2 r ) 2 = C 2 , ( t ) 2 = D 2 , ( q 2 ) 2 = E 2 , ( s 2 r ) 2 = F 2 {\displaystyle \left(y^{3}+{q \over 2}\right)^{2}=A^{2},({\sqrt {p}}y^{2})^{2}=B^{2},\left({\sqrt {r}}y+{s \over 2{\sqrt {r}}}\right)^{2}=C^{2},({\sqrt {t}})^{2}=D^{2},\left({q \over 2}\right)^{2}=E^{2},\left({s \over 2{\sqrt {r}}}\right)^{2}=F^{2}} ∴ A 2 + B 2 + C 2 + D 2 − E 2 − F 2 = 0 {\displaystyle \therefore \;\;\;A^{2}+B^{2}+C^{2}+D^{2}-E^{2}-F^{2}=0} A 2 + B 2 + C 2 = − D 2 + E 2 + F 2 {\displaystyle A^{2}+B^{2}+C^{2}=-D^{2}+E^{2}+F^{2}} {\displaystyle } 
![{\displaystyle {\sqrt[{3}]{{{1} \over {2}}+{{1} \over {6}}{\sqrt {{23} \over {3}}}}}+{\sqrt[{3}]{{{1} \over {2}}-{{1} \over {6}}{\sqrt {{23} \over {3}}}}}=1.324717957244746025960908854...}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c28ad1f1bd3bf034ab2430cbbda9c3752c1b1f12)
근과 계수의 관계 문서를 참고하십시오.
