Hi
General cubic equation \(\displaystyle ax^3+bx^2+cx+d=0\)
I have this cubic equation
\(\displaystyle x^3-y^2x+\dfrac{mn(m^2-n^2)}{y}=0\)
Not knowing how to go about solving for x I looked at wikipedia (yes, I know
)
http://en.wikipedia.org/wiki/Cubic_function#Special_cases
In the above I have \(\displaystyle a=1\ \ \ \ b=0\ \ \ \ c=-y^2\ \ \ \ d=\dfrac{mn(m^2-n^2)}{y}\)
Going through the part for 'Reduction to a depressed cubic' I thought the following was correct...
They set \(\displaystyle x=t-\dfrac{b}{3a}\)
since I have b=0 this would be x=t
Theirs \(\displaystyle t^2+pt+q=0\)
where \(\displaystyle p=\dfrac{3ac-b^2}{3a^2}\)
and \(\displaystyle q=\dfrac{2b^3-9abc+27a^2d}{27a^3}\)
again, since I have a=1 and b=0 these can be reduced to p=c and q=d (I think this is correct)
Moving down the page to the solution
\(\displaystyle t_1=\sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}+\sqrt[3]{-\dfrac{q}{2}-\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}\)
since I have p=c and q=d I substitute my values into the above...
\(\displaystyle x=\sqrt[3]{-\dfrac{\dfrac{mn(m^2-n^2)}{y}}{2}+\sqrt{\dfrac{\left(\dfrac{mn(m^2-n^2)}{y}\right)^2}{4}+\dfrac{(-y^2)^3}{27}}}+\sqrt[3]{-\dfrac{\dfrac{mn(m^2-n^2)}{y}}{2}-\sqrt{\dfrac{\left(\dfrac{mn(m^2-n^2)}{y}\right)^2}{4}+\dfrac{(-y^2)^3}{27}}}\)
I know that one solution of the original equation is
\(\displaystyle when\ m=77,\ \ \ \ n=38\ \ \ \ and\ \ \ \ y=78\ \ \ \ then\ \ \ \ x=55\)
I don't get this result when I plug m, n and y into the \(\displaystyle x=...\) equation above, so I'm obviously doing something wrong.
Any help would be welcome
Thanks
General cubic equation \(\displaystyle ax^3+bx^2+cx+d=0\)
I have this cubic equation
\(\displaystyle x^3-y^2x+\dfrac{mn(m^2-n^2)}{y}=0\)
Not knowing how to go about solving for x I looked at wikipedia (yes, I know
http://en.wikipedia.org/wiki/Cubic_function#Special_cases
In the above I have \(\displaystyle a=1\ \ \ \ b=0\ \ \ \ c=-y^2\ \ \ \ d=\dfrac{mn(m^2-n^2)}{y}\)
Going through the part for 'Reduction to a depressed cubic' I thought the following was correct...
They set \(\displaystyle x=t-\dfrac{b}{3a}\)
since I have b=0 this would be x=t
Theirs \(\displaystyle t^2+pt+q=0\)
where \(\displaystyle p=\dfrac{3ac-b^2}{3a^2}\)
and \(\displaystyle q=\dfrac{2b^3-9abc+27a^2d}{27a^3}\)
again, since I have a=1 and b=0 these can be reduced to p=c and q=d (I think this is correct)
Moving down the page to the solution
\(\displaystyle t_1=\sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}+\sqrt[3]{-\dfrac{q}{2}-\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}\)
since I have p=c and q=d I substitute my values into the above...
\(\displaystyle x=\sqrt[3]{-\dfrac{\dfrac{mn(m^2-n^2)}{y}}{2}+\sqrt{\dfrac{\left(\dfrac{mn(m^2-n^2)}{y}\right)^2}{4}+\dfrac{(-y^2)^3}{27}}}+\sqrt[3]{-\dfrac{\dfrac{mn(m^2-n^2)}{y}}{2}-\sqrt{\dfrac{\left(\dfrac{mn(m^2-n^2)}{y}\right)^2}{4}+\dfrac{(-y^2)^3}{27}}}\)
I know that one solution of the original equation is
\(\displaystyle when\ m=77,\ \ \ \ n=38\ \ \ \ and\ \ \ \ y=78\ \ \ \ then\ \ \ \ x=55\)
I don't get this result when I plug m, n and y into the \(\displaystyle x=...\) equation above, so I'm obviously doing something wrong.
Any help would be welcome
Thanks