# Where is the mistake?

Given that $$p$$ is a real number , then for what values of $$p$$ the equation ${(x+3p+1)}^{\frac{1}{3}} - x^ \frac{1}{3} = 1$ has real solutions.

${(x+3p+1)}^{\frac{1}{3}} - x^ \frac{1}{3} - 1 = 0$ if $$a+b+c = 0$$ we have $$a^3+b^3+c^3 = 3abc$$ $x+ 3p + 1 -x -1 = 3[(x+3p+1)(x)]^\frac{1}{3}$ $3p = 3[(x+3p+1)(x)]^\frac{1}{3}$ $p^3 = x(x + 3p +1)$ $x^2 + (3p + 1)x -p^3 =0$ For real roots $D \geq 0$ $4p^3 + 9p^2 + 6p +1 \geq 0$ $(p+1)^2(4p+1) \geq 0$ therefore , $p \geq \frac{-1}{4} \cup -1.$

Please help me understand which step did I do wrong? And what is the correct solution for the above question?

Note by Starwar Clone
2 years, 3 months ago

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Now, note that

$a+b+c=0 \implies a^3 + b^3 + c^3 = 3abc$

But, $a^3 + b^3 + c^3 = 3abc \not \Rightarrow a+b+c=0$

So, $$a+b+c=0$$ is not equivalent to $$a^3 + b^3 + c^3 = 3abc$$.

A better approach to solve this is putting $$x=y^3 \iff x^{\frac{1}{3}} = y$$.

Then, the equation becomes:

\begin{align}(y^3 + 3p +1)^{\frac{1}{3}} &= 1+y \\\iff y^3 + 3p +1 &= (1+y)^3 \\\iff y^2 + y -p &= 0 \end{align}

So, for the roots of the original equation to be real, we need the discriminant of $$y^2 + y - p=0$$ to be non negative.

That is, $1+4p \geq 0 \\\iff p \geq -\frac{1}{4}$

- 2 years, 3 months ago

I got what you meant here but still i think the answer is wrong , check on wolframalpha putting p = -0.25 , it says no solution exists to the below equation for ${(x+0.25)}^{\frac{1}{3}} - x^ \frac{1}{3} - 1 = 0$

- 2 years, 3 months ago

WolframAlpha gives you the option to "use the real-valued root instead", (in blue just below the entry box), which in this case will give you a value of $$x = -\frac{1}{8}$$ as Deeparaj has pointed out.

- 2 years, 3 months ago

- 2 years, 3 months ago

Oh yes , i was working only with principal root value and hence was getting no solution

- 2 years, 3 months ago

Try $$x = \frac{-1}{8}$$

- 2 years, 3 months ago

What does $$\frac{-1}{4} \cup 1$$ mean?

Staff - 2 years, 3 months ago

i mean that p>= -0.25 and also p = -1 is a solutionof the inequality
$(p+1)^2(4p+1) \geq 0$

- 2 years, 3 months ago

@Brian Charlesworth @Jon Haussmann , your help will be valuable .

- 2 years, 3 months ago

I used the same approach as Deeparaj, but I think we need more than just the discriminant to be non-negative; in order to avoid complex values for $$y = x^{\frac{1}{3}}$$ we will require that $$x \ge 0$$ and thus also $$y \ge 0$$. So we require that

$$y = \dfrac{-1 + \sqrt{1 + 4p}}{2} \ge 0 \Longrightarrow \sqrt{1 + 4p} \ge 1 \Longrightarrow p \ge 0$$,

which seems to be in agreement with results from WolframAlpha.

- 2 years, 3 months ago

Don't we define $$x^{\frac{1}{3}} = -(-x)^{\frac{1}{3}} \quad x < 0$$ right?

- 2 years, 3 months ago

Yes, we could choose to work with the real-valued root, in which case WolframAlpha does return a value of $$x = -\frac{1}{8}$$ for $$p = -\frac{1}{4}$$, and no solutions if $$p \lt -\frac{1}{4}$$.

(WolframAlpha can be a bit difficult when taking fractional roots of negative numbers.)

- 2 years, 3 months ago

this is exactly what i was confused about . Thank you for helping me out with the wolframalpha query .

- 2 years, 3 months ago