# Kyuubi-c?-1

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 $$a+b+c$$.

Note by Ayush G Rai
2 years, 1 month ago

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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: $a^3-1=(a-1)(a^2+a+1)\\\implies \frac{1}{a-1}=\frac{a^2+a+1}{a^3-1}$

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

- 2 years, 1 month ago

awesome approach .! +1.!

- 2 years, 1 month ago