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 $a+b+c$ .

#Algebra

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
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`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

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

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

A Former Brilliant Member - 5 years ago

awesome approach .! +1.!

Rishabh Tiwari - 5 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$