- Posmatrajmo polinom trećeg stepena
za ![{\displaystyle a\neq 0}](//wikimedia.org/api/rest_v1/media/math/render/svg/f455a7f96d74aa94573d8e32da3b240ab0aa294f)
- koja siječe x-osu u tačkama koje su nule odgovarajuće jednačine trećeg stepena (kubne jednačine).
![{\displaystyle ax^{3}+bx^{2}+cx+d=0}](//wikimedia.org/api/rest_v1/media/math/render/svg/1c6a382654ad8c94dc3bfea84bf4a869ab5c68cb)
- Često se diskriminantom kubne jednačine naziva diskriminanta [2]
![{\displaystyle D=a^{4}(x_{1}-x_{2})^{2}(x_{2}-x_{3})^{2}(x_{3}-x_{1})^{2})}](//wikimedia.org/api/rest_v1/media/math/render/svg/28f3327a1b62b54b8d54818b505fb657fac00c53)
- pripadnog polinoma :
gdje su
korijeni polinoma (rješenja date jednačine. Vrijedi
![{\displaystyle D=-4b^{3}d+b^{2}c^{2}-4ac^{3}+18abcd-27a^{2}d^{2}.}](//wikimedia.org/api/rest_v1/media/math/render/svg/5a1a2c1735ae5203fc901def8c25de17134f12ba)
- Ovo vrijedi za sve kubne jednačine, a ne samo za one s realnim koeficijentima.
- I izraz
je diskriminanta
- Za
ima jedno realno, dva konjugovano kompleksna rješenja
- Za
ima sva tri realna, bar jedno dvostruko rješenje
- Za
ima sva tri realna i različita rješenja
- Primjer
Jednačina
smjenom
![{\displaystyle y=x+1}](//wikimedia.org/api/rest_v1/media/math/render/svg/14d4839f41dc3d34ae434dfad1ab18479cc6a41b)
- postaje
![{\displaystyle y^{3}-4y=0}](//wikimedia.org/api/rest_v1/media/math/render/svg/5e08547db90a9abca17362daacb11a0e916a4b30)
- Jednačina
smjenom
![{\displaystyle y=u+v}](//wikimedia.org/api/rest_v1/media/math/render/svg/07fa8911b30ee68a25e5882fd01e641fa16915e2)
postaje
![{\displaystyle u^{3}+v^{3}+\left(3uv+p\right)\left(u+v\right)+q=0}](//wikimedia.org/api/rest_v1/media/math/render/svg/db02a952114ab0715e4093e263f0f506486a01f2)
- Ovdje smo nepoznatu
zamijenili sa 2 nove
i
.
uvodimo novi uslov
![{\displaystyle 3uv=p}](//wikimedia.org/api/rest_v1/media/math/render/svg/b2b4d280e4762bc0eca22990661ca9e273bebf30)
- pa je
![{\displaystyle u^{3}+v^{3}=-q;uv={\frac {-p}{3}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/9c7cfd4627dd04a1b3f97b6eaca2ee42aee7264c)
- Iz
![{\displaystyle y^{3}-4y=0}](//wikimedia.org/api/rest_v1/media/math/render/svg/5e08547db90a9abca17362daacb11a0e916a4b30)
- dobijamo
![{\displaystyle u^{3}+v^{3}=0;uv={\frac {4}{3}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/f072a2cc7d31946f36fdb6d2bf94eb564ad42ff8)
- Sistem
ekvivlentan je sa
![{\displaystyle u^{3}+v^{3}=-q;u^{3}v^{3}={\frac {-p^{3}}{27}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/a835e8eacf34e65b46b3eceeff59231f2752c10d)
- Iz kojeg dobijamo
![{\displaystyle t^{2}+qt-{\frac {-p^{3}}{27}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/0ac0721aecb095fd4c9c56a41178cb6940691964)
- za
![{\displaystyle t_{1}=u^{3}it_{2}=v^{3}}](//wikimedia.org/api/rest_v1/media/math/render/svg/0e8f0ab2d3b2936caf305ac461e4893c8ec35b0a)
- Njenim rješavanjem dobijamo
![{\displaystyle t_{1},2=-{\frac {q}{27}}\pm {\sqrt {({\frac {q^{2}}{4}})+{\frac {p^{3}}{27}}}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/53a079011b5bdb2b180ce1561c849de35d639f64)
- Vračanjem druge smjene dobijamo
![{\displaystyle y={\sqrt[{3}]{{\frac {-q}{2}}+{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}+{\sqrt[{3}]{{\frac {-q}{2}}-{\sqrt {{\frac {q^{2}}{4}}+{\frac {p^{3}}{27}}}}}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/e3ec61fe60abed23b474dc87be6605fb30e9dbb9)
- IZ
![{\displaystyle y^{3}-4y=0}](//wikimedia.org/api/rest_v1/media/math/render/svg/5e08547db90a9abca17362daacb11a0e916a4b30)
![{\displaystyle p=-4iq=0}](//wikimedia.org/api/rest_v1/media/math/render/svg/c72323caf21675415ad098db1b5d0ea1024925be)
- pa je
![{\displaystyle y=0}](//wikimedia.org/api/rest_v1/media/math/render/svg/094f824655138f6b11d96a0da32e7f0716ba6959)
- tj. jedno rješenje jednačine je
![{\displaystyle y=x^{3}-3x^{2}-x+3}](//wikimedia.org/api/rest_v1/media/math/render/svg/3e915776e9703fda3db058e60c2b715c3c7e02ad)
![{\displaystyle x=o}](//wikimedia.org/api/rest_v1/media/math/render/svg/bba0e2c15285c6265d7ebe44fbaeda2ff6826935)
- Parovi (u,v) su rješenja sistema
![{\displaystyle u^{3}+v^{3}=-q;uv={\frac {-p}{3}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/9c7cfd4627dd04a1b3f97b6eaca2ee42aee7264c)
- Iz jednačina
![{\displaystyle u^{3}=t_{1}iv^{3}=t_{2}}](//wikimedia.org/api/rest_v1/media/math/render/svg/21276d9d4dfe88b719edf33a90f918c243c4a952)
- slijedi
,
,
![{\displaystyle u_{3}=a^{2}{\sqrt[{3}]{t}}_{1}}](//wikimedia.org/api/rest_v1/media/math/render/svg/af023ac3fdacc843cb206b2fc1edad780d2346a9)
,
,
![{\displaystyle v_{3}=a^{2}{\sqrt[{3}]{t}}_{2}}](//wikimedia.org/api/rest_v1/media/math/render/svg/c1b23ae32e1dc56c905fc0685499f3ded1c5e01f)
- gdje su treči korjeni jedinice
![{\displaystyle a^{0}=1,a^{1}={\frac {1}{2}}(-1+i{\sqrt[{3}]{3}}),a^{2}={\frac {1}{2}}(-1-i{\sqrt[{3}]{3}})}](//wikimedia.org/api/rest_v1/media/math/render/svg/7ca3baa450e765cb4f8bb2f990d8a395837d1af1)
Odnosno sva rješenja jednačine
![{\displaystyle y^{3}+py+q=0}](//wikimedia.org/api/rest_v1/media/math/render/svg/c5f60f6e3e2fd36f89d47649d68591331b66cf2f)
- data su formulom
,
,
![{\displaystyle y_{3}=a^{2}{\sqrt[{3}]{t}}_{1}+{\sqrt[{3}]{t}}_{2}}](//wikimedia.org/api/rest_v1/media/math/render/svg/6321eb187df7593e5ae90a16e3ffafd110b2abc2)
- koje se također nazivaju Kardanove formule.
- U našem slučaju (za
) rješenja jednačine su realna, ali se do njih dolazi izračunavanjem kubnih korjena imaginarnih brojeva. Zato što se tada ne možemo osloboditi imaginarnosti u Kardanovim formulama, ovaj slučaj
) nazivamo nesvodljiv slučaj.
- Za
i
imamo
![{\displaystyle D={\frac {-64}{27}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/80de0302a0b6fe3e7aa23be405ac652d02bb278e)
- pa je
i imamo 2 slučaja
, ![{\displaystyle y_{3}=a^{2}{\sqrt[{3}]{t}}_{1}+{\sqrt[{3}]{t}}_{2}}](//wikimedia.org/api/rest_v1/media/math/render/svg/6321eb187df7593e5ae90a16e3ffafd110b2abc2)
- Iz ovoga je
![{\displaystyle t_{1}=-t_{2}}](//wikimedia.org/api/rest_v1/media/math/render/svg/b2eabbea28d2926f0174bfc96a662b698a16c9f7)
![{\displaystyle y_{2}=(a-a^{2}){\sqrt[{3}]{\sqrt {\frac {-64}{27}}}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/abaa9562650b193691241066217f74e4d60aa2b3)
=>
![{\displaystyle y_{2}=(i{\sqrt[{3}]{3}}){\frac {2i}{\sqrt[{3}]{3}}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/5b7707004262d4eafa6a9b6217e2e20039bbf769)
=>
![{\displaystyle y_{2}=2\land y_{3}=-2}](//wikimedia.org/api/rest_v1/media/math/render/svg/78716a6cf83e7f57cd4e4402a8947904813a4bb1)
pa je rješenje polaznog primjera
![{\displaystyle x_{1}=-1}](//wikimedia.org/api/rest_v1/media/math/render/svg/dcd2c509bbdf4efec956cffd6946a3e0d2db8cb1)
![{\displaystyle x_{2}=1,}](//wikimedia.org/api/rest_v1/media/math/render/svg/3e13651ca80d3aa6e15f3a4e59e740449b35ea39)
![{\displaystyle x_{3}=3}](//wikimedia.org/api/rest_v1/media/math/render/svg/7642462aea531460450aa2d88e1388587cbdf3d5)
Opće rješenje za svaku kubnu jednačinu
![{\displaystyle ax^{3}+bx^{2}+cx+d=\,0}](//wikimedia.org/api/rest_v1/media/math/render/svg/f148d5788a335b7fdb2c96c908e9e807618f6b4d)
- je
![{\displaystyle {\begin{aligned}x_{1}=&-{\frac {b}{3a}}\\&-{\frac {1}{3a}}{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d+{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\\&-{\frac {1}{3a}}{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\\x_{2}=&-{\frac {b}{3a}}\\&+{\frac {1+i{\sqrt {3}}}{6a}}{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d+{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\\&+{\frac {1-i{\sqrt {3}}}{6a}}{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\\x_{3}=&-{\frac {b}{3a}}\\&+{\frac {1-i{\sqrt {3}}}{6a}}{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d+{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\\&+{\frac {1+i{\sqrt {3}}}{6a}}{\sqrt[{3}]{\frac {2b^{3}-9abc+27a^{2}d-{\sqrt {\left(2b^{3}-9abc+27a^{2}d\right)^{2}-4\left(b^{2}-3ac\right)^{3}}}}{2}}}\end{aligned}}}](//wikimedia.org/api/rest_v1/media/math/render/svg/ef8fdcb363d20258b4ec96f33fc248aae912c9ff)