The real root of the equation \(8x^3-3x^2-3x-1=0\) can be written in the form \(\dfrac{\sqrt [3]{a}+\sqrt [3]{b}+1}{c},\) where \(a,b\) and \(c\) are positive integers. Find \(a+b+c\).

Using the expansion of \((1+x)^3\) we get the following:
\[\begin{align} 9x^3&=(1+x)^3 \\\implies \sqrt[3]{9}x&=1+x \\\implies x&=\frac{1}{\sqrt[3]{9}-1} \\\implies x&=\frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}\end{align}\]

The last step follows from the following identity: \[a^3-1=(a-1)(a^2+a+1)\\\implies \frac{1}{a-1}=\frac{a^2+a+1}{a^3-1}\]

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TopNewestUsing the expansion of \((1+x)^3\) we get the following: \[\begin{align} 9x^3&=(1+x)^3 \\\implies \sqrt[3]{9}x&=1+x \\\implies x&=\frac{1}{\sqrt[3]{9}-1} \\\implies x&=\frac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}\end{align}\]

The last step follows from the following identity: \[a^3-1=(a-1)(a^2+a+1)\\\implies \frac{1}{a-1}=\frac{a^2+a+1}{a^3-1}\]

Hence, \(a+b+c=\boxed{98}\)

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awesome approach .! +1.!

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