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.

My answer:

\[ {(x+3p+1)}^{\frac{1}{3}} - x^ \frac{1}{3} - 1 = 0 \] 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?

## Comments

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TopNewestNow, 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 equivalentto \(a^3 + b^3 + c^3 = 3abc\).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, \[ 1+4p \geq 0 \\\iff p \geq -\frac{1}{4} \] – Deeparaj Bhat · 11 months, 1 week ago

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– Starwar Clone · 11 months, 1 week 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 \[ {(x+0.25)}^{\frac{1}{3}} - x^ \frac{1}{3} - 1 = 0 \]Log in to reply

– Brian Charlesworth · 11 months, 1 week 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.Log in to reply

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– Starwar Clone · 11 months, 1 week ago

Oh yes , i was working only with principal root value and hence was getting no solutionLog in to reply

– Deeparaj Bhat · 11 months, 1 week ago

Try \(x = \frac{-1}{8} \)Log in to reply

What does \(\frac{-1}{4} \cup 1\) mean? – Agnishom Chattopadhyay · 11 months, 1 week ago

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\[ (p+1)^2(4p+1) \geq 0\] – Starwar Clone · 11 months, 1 week ago

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@Brian Charlesworth @Jon Haussmann , your help will be valuable . – Starwar Clone · 11 months, 1 week ago

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

which seems to be in agreement with results from WolframAlpha. – Brian Charlesworth · 11 months, 1 week ago

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– Deeparaj Bhat · 11 months, 1 week ago

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

(WolframAlpha can be a bit difficult when taking fractional roots of negative numbers.) – Brian Charlesworth · 11 months, 1 week ago

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– Starwar Clone · 11 months, 1 week ago

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