Real Root

Algebra Level 4

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 $a+b+c$ .

The answer is 98.

2 solutions

Danish Ahmed
Apr 25, 2015

We have that

$9x^3-x^3-3x^2-3x-1=0$

$9x^3 = (x+1)^3$ ,

so it follows that

$x\sqrt[3]{9} = x+1$ .

Solving for $x$ yields $\dfrac{1}{\sqrt[3]{9}-1} = \dfrac{\sqrt[3]{81}+\sqrt[3]{9}+1}{8}$ ,

so the answer is $\boxed{98}$ .

Moderator note:

Great work!

It has already been shown:

If two real numbers have the same cube then they are equal. Then we are left with a linear equation with one solution. So it has only one real solution.

Joel Tan - 6 years, 1 month ago

I assumed $b=a^2$ and did it by trial and error.

Akiva Weinberger - 6 years, 1 month ago

This is very good for this specific question. However, general way is shown by another solver.

Lu Chee Ket - 5 years, 8 months ago

Exactly same

Department 8 - 5 years, 8 months ago

Nice solution!! (+1). I did it in a bit lengthy process, using Cardano's method you know.... However, this was really nice

Raushan Sharma - 5 years, 2 months ago

Otto Bretscher
Apr 30, 2015

Danish's solution is great! Just for the sake of variety (and for practice), let's do it with the cubic formula; it's a pretty straightforward computation.

Make the substitution $y=2x-\frac{1}{4}$ to eliminate the quadratic term; now we have $y^3-\frac{27}{16}y-\frac{45}{32}=0$ . The cubic formula gives a single real solution, $y=\sqrt[3]{\frac{45}{64}+\frac{9}{16}}+\sqrt[3]{\frac{45}{64}-\frac{9}{16}}=\frac{1}{4}\left(\sqrt[3]{81}+\sqrt[3]{9}\right).$ Now $x=\frac{1}{8}+\frac{y}{2}=\frac{1+\sqrt[3]{81}+\sqrt[3]{9}}{8}$ .

Real Root

The answer is 98.

2 solutions

Moderator note:

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