Sum of roots of $p-3 = 5p^{2/3} - 6p^{1/3}=0$?

Question

In the middle of a solution of a question, the equation $$p-3 = 5p^{2/3} - 6p^{1/3}=0$$ is given. I am required to find out the value of the sum of roots for the next step, which for an equation $ax^3+bx^2+cx+d$ is equal to $-b/a$. I tried to put the equation in this format by cubing both sides, however I get $$p^3-27-9p^2+27p = 125p^2 - 216p - 90p(5p^{2/3} - 6p^{1/3})$$ which is in terms of $p^{2/3}$, therefore I think it cannot be right. What should I do? Is there any other method, or have I missed something in this method?

Answer 1

If $$p^3-27-9p^2+27p = 125p^2 - 216p - 90p(5p^{2/3} - 6p^{1/3})$$ (I haven't checked that your algebra is correct) you can substitute $$5p^{2/3} - 6p^{1/3}=p-3$$ to get rid of the fractional powers.

Note. The pair of equations $p-3 = 5p^{2/3} - 6p^{1/3}=0$ has no solution, I am assuming you meant the single equation $p-3 = 5p^{2/3} - 6p^{1/3}$.

Answer 2

Hint: $p^{2/3} = (p^{1/3})^2$

Answer 3

I do believe that this could be a bit simpler if you observe that the equation itself is already cubic by nature with variable $p^{1/3}$.

$$ \left(p^{1/3}\right)^3−3=5\left(p^{1/3}\right)^2−6p^{1/3} $$

So, assume that $q=p^{1/3}$, you can rewrite this cubic as

$$q^3 - 5q^2+6q-3=0$$

Assume this has roots $r$, $s$ and $t$, then we know from Vieta's formulas that :

• $r+s+t=5$
• $rs+st+tr=6$
• $rst=3$

We are actually interested in $r^3+s^3+t^3$ which we can obtain as :

$$r^3+s^3+t^3 = (r+s+t)^3 - 3 (r+s+t)(rs+st+tr)+3rst $$

and this leads to the final answer : $5^3 - 3\cdot5\cdot6+3\cdot3 = 44 $