Question

Find the conditions that $a{x}^3+b{x}^2+cx+d$ may be a perfect cube.

Answer 1

Let $a{x}^3+b{x}^2+cx+d$ be the cubic equation having roots $\alpha .\beta ,\gamma$

As it is perfect cube this means that , all of its roots are equal ,

$\alpha =p$

$\beta =p$

$\gamma =p$

Sum of the roots of the cubic equation wil be $\alpha +\beta +\gamma =-\frac{x^2\left(coefficient\right)}{x^3\left(coefficient\right)}$

$\alpha +\beta +\gamma =3p=-\frac{b}{a}$

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{x\left(coefficient\right)}{x^3\left(coefficient\right)}$

$p^2+p^2+p^2=\frac{c}{a}$

$3p^2=\frac{c}{a}$

Product of the roots of the cubic equation will be $\alpha \beta \gamma =p^3=-\frac{constant}{x^3\left(coefficient\right)}$

$p^3=-\frac{d}{a}$

$3p=-\frac{b}{a}$

cubing both side we get .

$27p^3=-\frac{b^3}{a^3}$

putting the value of $p^3$ from the product of the equation and thus we get ,

$-27\frac{d}{a}=-\frac{b^3}{a^3}$

$b^3=27d{a}^2$

Similarly ,

$3p^2=\frac{c}{a}$

cubing both side we get,

$27p^6=\frac{c^3}{a^3}$

$27{\left(p^3\right)}^2=\frac{c^3}{a^3}$

putting the value of $p^3$ from the product of the equation and thus we get ,

$27\frac{d^2}{a^2}=\frac{c^3}{a^3}$

$c^3=27a{d}^2$

Thus the condition for the perfect cube for the cubic equation $a{x}^3+b{x}^2+cx+d$ will be $b^3=27d{a}^2$ and $c^3=27a{d}^2$