The equation \begin{align*}
9x³-3x²-3x-1 & =0
\end{align*} has a real root of the form $\frac{√3a+√3b+1}{c}$
where
a, b, c are positive integers. Find a + b + c.
The equation \begin{align*}
9x³-3x²-3x-1 & =0
\end{align*} has a real root of the form $\frac{√3a+√3b+1}{c}$
where
a, b, c are positive integers. Find a + b + c.
I think you didn't type the question correctly.
Short solution:
Long solution for which the question isn't typed correctly:
$10x^3-(x+1)^3=0 \cdots (1)$
Or,$(x\sqrt[3]{10})^3-(x+1)^3=0$
Or,$(x\sqrt[3]{10}-x-1)(x^2\sqrt[3]{100}+x^2\sqrt[3]{10}+x\sqrt[3]{10}+x+1)=0$
Either,
$(x\sqrt[3]{10}-x-1)=0$
This will lead to the soluton above
Or,
$(x^2\sqrt[3]{100}+x^2\sqrt[3]{10}+x\sqrt[3]{10}+x+1)=0$
$x^2(\sqrt[3]{100}+\sqrt[3]{10})+x(\sqrt[3]{10}+1)+1=0$
Now the discriminant is,
$(\sqrt[3]{10}+1)2-4(\sqrt[3]{100}+\sqrt[3]{10})$
$=\sqrt[3]{100}+2\sqrt[3]{10}+1-4\sqrt[3]{100}-4\sqrt[3]{10}$
$=1-3\sqrt[3]{100}-2\sqrt[3]{10}<0$
So there are no other real roots.