Question

Let $a,b,c,p$ be rational numbers such that $p$ is not a perfect cube.
$a+b{p}^{\frac{1}{3}}+c{p}^{\frac{2}{3}}=0$, then prove that $a=b=c=0$

Answer 1

We have,
$a+b{p}^{\frac{1}{3}}+c{p}^{\frac{2}{3}}=0$
Multiplying both sides by $p^{\frac{1}{3}}$, we get
$a{p}^{\frac{1}{3}}+b{p}^{\frac{2}{3}}+cp=0$
Multiplying $\left(i\right)$by $b$ and $\left(ii\right)$ by $c$ and subtracting, we get
$(ab+b^2{p}^{\frac{1}{3}}+bc{p}^{\frac{2}{3}})-(ac{p}^{\frac{1}{3}}+bc{p}^{\frac{2}{3}}+c^{2}p)=0$
$\Rightarrow (b^2-ac)p^{\frac{1}{3}}+ab-c^{2}p=0$
${\Rightarrow b}^2-ac=0$ and $ab-c^{2}p=0$
$\Rightarrow b^2=ac$ and $ab=c^{2}p$
$\Rightarrow b^2=ac$ and $a^2{b}^2=c^4{p}^2$
$\Rightarrow a^2\left(ac\right)=c^4{p}^2$
${\Rightarrow a}^{3}c-p^2{c}^4=0$
$\Rightarrow (a^3-p^2{c}^3)c=0$
$\Rightarrow a^3-p^2{c}^3=0$ or $c=0$
Now,
$a^3-p^2{c}^3=0$
$p^2=\frac{a^3}{c^3}\Rightarrow {\left(p^2\right)}^{\frac{1}{3}}={\left(\frac{a^3}{c^3}\right)}^{\frac{1}{3}}\Rightarrow {\left(p^{\frac{1}{3}}\right)}^2={\left\{{\left(\frac{a}{c}\right)}^3\right\}}^{\frac{1}{3}}\Rightarrow {\left(p^{\frac{1}{3}}\right)}^2=\frac{a}{c}$
This is not possible as $p^{1/3}$ is irrational and $\frac{a}{c}$ is rational.
$\therefore a^3-p^2{c}^3\ne 0$
Hence $c=0$
Putting $c=0$ in $b^2-ac=0$, we get $b=0$
Putting $b=0$ and $c=0$ in $a+b{p}^{\frac{1}{3}}+c{p}^{\frac{2}{3}}=0$ we get $a=0$
Hence $a=b=c=0$