Qubic Formula

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} & \text{Standard form of Qubic equation} \\ & \text{ }a{{x}^{3}}+b{{x}^{2}}+cx+d=0\text{ (where }a\ne \text{0)} \\ & \text{Let} \\ & \text{ }x=y-\frac{b}{3a} \\ & \text{Then} \\ & \text{ a}{{\left( y-\frac{b}{3a} \right)}^{3}}+b{{\left( y-\frac{b}{3a} \right)}^{2}}+c\left( y-\frac{b}{3a} \right)+d=0 \\ & \text{Solving this,} \\ & \text{We get,} \\ & \text{ }a{{y}^{3}}+\left( \frac{3ac-{{b}^{2}}}{3a} \right)y+\left( \frac{2{{b}^{3}}-9abc+27{{a}^{2}}d}{27{{a}^{2}}} \right) \\ & \text{Here} \\ & \text{Put,} \\ & \text{ }\frac{3ac-{{b}^{2}}}{3a}=B\text{ AND }\frac{2{{b}^{3}}-9abc+27{{a}^{2}}d}{27{{a}^{2}}}=C \\ & \text{We get} \\ & \text{ }a{{y}^{3}}+By+C \\ & \text{ }a{{y}^{3}}+By+C=0 \\ & \text{ Dividing }a\text{ on both sides} \\ & \text{ }{{y}^{3}}+\frac{B}{a}y+\frac{C}{a}=0.............(i) \\ & \text{Let } \\ & \text{ }y=u+v.............(iv) \\ & \text{Taking cube on both sides} \\ & \text{ }{{\left( y \right)}^{3}}={{\left( u+v \right)}^{3}} \\ & \text{we get } \\ & \text{ }{{y}^{3}}={{u}^{3}}+{{v}^{3}}+3uv(u+v) \\ & \text{So,} \\ & \text{ }{{y}^{3}}={{u}^{3}}+{{v}^{3}}+3uvy \\ & \text{ }{{y}^{3}}-3uvy-\left( {{u}^{3}}+{{v}^{3}} \right)=0.............(ii) \\ & \text{Comparing (i) and (ii), we get} \\ & 3uv=-\frac{B}{a}\text{ and }{{u}^{3}}+{{v}^{3}}=\frac{C}{a} \\ & {{\left( uv \right)}^{3}}={{\left( -\frac{B}{3a} \right)}^{3}}\text{ and\text{ }}-\left( {{u}^{3}}+{{v}^{3}} \right)=\frac{C}{a} \\ & {{u}^{3}}{{v}^{3}}=-\frac{B^3}{27{{a}^{3}}}\text{ and\text{ }}{{u}^{3}}+{{v}^{3}}=-\frac{C}{a} \\ & \text{Now,} \\ & \text{We will create a quadratic equation} \\ & \text{We know that } \\ & \text{ }{{y}^{2}}-Sy+P=0 \\ & \text{So,} \\ & \text{ }{{y}^{2}}-\left( -\frac{C}{a} \right)y+\left( -\frac{B^3}{27{{a}^{3}}} \right)=0 \\ & \text{ }{{y}^{2}}+\left( \frac{C}{a} \right)y+\left( -\frac{B^3}{27{{a}^{3}}} \right)=0 \\ & \text{Here,} \\ & \text{ }a=1\text{ }b=\frac{C}{a}\text{ and }c=-\frac{B^3}{27{{a}^{3}}} \\ & \text{According to Quadratic formula,} \\ & \text{ }y=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} \\ & \text{So,} \\ & \text{ }y=\frac{-\left( \frac{C}{a} \right)\pm \sqrt{{{\left( \frac{C}{a} \right)}^{2}}-4(1)\left( -\frac{B^3}{27{{a}^{3}}} \right)}}{2(1)}.............(iii) \\ & \text{Multiplying }\frac{a}{a}=1\text{ with equation (iii)} \\ & \text{ }y=\left( \frac{a}{a} \right)\left( \frac{-\left( \frac{C}{a} \right)\pm \sqrt{{{\left( \frac{C}{a} \right)}^{2}}-4(1)\left( -\frac{B^3}{27{{a}^{3}}} \right)}}{2(1)} \right) \\ & \text{ }y=\frac{-C\pm \sqrt{{{C}^{2}}+\frac{4B^3}{27a}}}{2a} \\ & \text{Here,} \\ & \text{ }{{u}^{3}}=\frac{-C+\sqrt{{{C}^{2}}+\frac{4B^3}{27a}}}{2a}\text{ and }{{v}^{3}}=\frac{-C-\sqrt{{{C}^{2}}+\frac{4B^3}{27a}}}{2a} \\ & \text{ }u=\sqrt[3]{\frac{-C+\sqrt{{{C}^{2}}+\frac{4B^3}{27a}}}{2a}}\text{ and }{{v}^{3}}=\sqrt[3]{\frac{-C-\sqrt{{{C}^{2}}+\frac{4B^3}{27a}}}{2a}} \\ & \text{From equ (iv),} \\ & \text{we get } \\ & \text{ }y=\sqrt[3]{\frac{-C+\sqrt{{{C}^{2}}+\frac{4B^3}{27a}}}{2a}}+\sqrt[3]{\frac{-C-\sqrt{{{C}^{2}}+\frac{4B^3}{27a}}}{2a}} \\ & \text{Also,} \\ & \text{ }x=y-\frac{b}{3a} \\ & \text{So,} \\ & \text{ x=}\left( \sqrt[3]{\frac{-C+\sqrt{{{C}^{2}}+\frac{4B^3}{27a}}}{2a}}+\sqrt[3]{\frac{-C-\sqrt{{{C}^{2}}+\frac{4B^3}{27a}}}{2a}} \right)-\frac{b}{3a} \\ & \text{Which is Qubic formula for Real root of Qubic equation} \\ & \text{Here,} \\ & \text{ }B=\frac{3ac-{{b}^{2}}}{3a}\text{ and }C=\frac{2{{b}^{3}}-9abc+27{{a}^{2}}d}{27{{a}^{2}}} \\ & \text{Note:-This equation will not work if} \\ & \text{ }\sqrt{{{C}^{2}}+\frac{4B^3}{27a}}\text{ is negative } \\ & \text{Standard form of Qubic equation} \\ & \text{ }a{{x}^{3}}+b{{x}^{2}}+cx+d=0\text{ (where }a\ne \text{0)} \\ & \text{Dividing }a\text{ on both sides} \\ & \text{ }{{x}^{3}}+\frac{b}{a}{{x}^{2}}+\frac{c}{a}x+\frac{d}{a}=0.............(i) \\ & \text{First root is} \\ & \text{ }x=\left( \sqrt[3]{\frac{-C+\sqrt{{{C}^{2}}+\frac{4B^3}{27a}}}{2a}}+\sqrt[3]{\frac{-C-\sqrt{{{C}^{2}}+\frac{4B^3}{27a}}}{2a}} \right)-\frac{b}{3a} \\ & \text{Let this root be t} \\ & \text{So, }3\text{ roots will be t, y and z } \\ & \text{We can write factorize it as} \\ & \text{ }\left( x-t \right)\left( x-y \right)\left( x-z \right) \\ & \text{Solving this we get} \\ & \text{ }{{x}^{3}}-\left( t+y+z \right){{x}^{2}}+\left( ty+yz+tz \right)x-tyz=0.............(ii)\text{ } \\ & \text{Comparing }(i)\text{ and }(ii),\text{ } \\ & \text{We get} \\ & \text{ }-\left( t+y+z \right)=\frac{b}{a}\text{ }-tyz=\frac{d}{a} \\ & \text{ }y+z=-\frac{b}{a}-t\text{ }yz=-\frac{d}{at}\text{ } \\ & \text{Now we create a Quadratic equation,} \\ & \text{We know that } \\ & \text{ }{{x}^{2}}-Sx+P=0 \\ & \text{So,} \\ & \text{ }{{x}^{2}}-\left( -\frac{b}{a}-t \right)x+\left( -\frac{d}{at} \right)=0 \\ & \text{Here,} \\ & \text{ a=1 b= }\frac{b}{a}+t\text{ and c=}-\frac{d}{at} \\ & \text{We know that,} \\ & \text{ }x=\frac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}\text{ } \\ & \text{So,} \\ & \text{ }x=\frac{-\left( \frac{b}{a}+t \right)\pm \sqrt{{{\left( \frac{b}{a}+t \right)}^{\text{2}}}+4\left( -\frac{d}{at} \right)}}{2} \\ & \text{Multiplying }\frac{a}{a}=1, \\ & \text{ }{{x}_{{}^{2}\!\!\diagup\!\!{}_{3}\;}}=\left( \frac{a}{a} \right)\left( \frac{-\left( \frac{b}{a}+t \right)\pm \sqrt{{{\left( \frac{b}{a}+t \right)}^{\text{2}}}+4\left( -\frac{d}{at} \right)}}{2} \right) \\ & \text{ }{{x}_{{}^{2}\!\!\diagup\!\!{}_{3}\;}}=\frac{-\left( b+at \right)\pm \sqrt{{{\left( b+at \right)}^{2}}+\frac{4ad}{t}}}{2a} \\ & \text{Above give two real/complex roots of Cubic equations} \\ & \text{where} \\ & \text{ t=}{{x}_{1}}=\left( \sqrt[3]{\frac{-C+\sqrt{{{C}^{2}}+\frac{4B^3}{27a}}}{2a}}+\sqrt[3]{\frac{-C-\sqrt{{{C}^{2}}+\frac{4B^3}{27a}}}{2a}} \right)-\frac{b}{3a} \\ & \text{Note:-The roots are complex or real it depends on value of }\sqrt{{{\left( b+at \right)}^{2}}+\frac{4ad}{t}} \\ & \text{If} \\ & \text{ (i) }\sqrt{{{\left( b+at \right)}^{2}}+\frac{4ad}{t}}\text{ is negative, two roots are complex} \\ & \text{ (ii) }\sqrt{{{\left( b+at \right)}^{2}}+\frac{4ad}{t}}\text{ is positive, two roots are real} \\ & \text{ } \\ & \text{ } \\ \end{align}}

References

These equations are created by Azan Shahid

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