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Question Number 80161 by peter frank last updated on 01/Feb/20

Find the relation between q and r so that x^3 +3px^2 +qx+r is a perfect cube for all value of x

$${Find}\:{the}\:{relation}\:{between} \\ $$$${q}\:{and}\:{r}\:\:{so}\:\:{that} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{px}^{\mathrm{2}} +{qx}+{r}\:{is}\:{a}\:{perfect} \\ $$$${cube}\:{for}\:{all}\:\:{value}\:{of}\:{x} \\ $$

Commented by peter frank last updated on 31/Jan/20

yes sir

$${yes}\:{sir} \\ $$

Commented by mr W last updated on 31/Jan/20

you mean x^3 +3px^2 +qx+r ?

$${you}\:{mean}\:{x}^{\mathrm{3}} +\mathrm{3}{px}^{\mathrm{2}} +{qx}+{r}\:? \\ $$

Commented by mr W last updated on 31/Jan/20

⇒q=3p^2 ⇒r=p^3 then x^3 +3px^2 +qx+r=(x+p)^3

$$\Rightarrow{q}=\mathrm{3}{p}^{\mathrm{2}} \\ $$$$\Rightarrow{r}={p}^{\mathrm{3}} \\ $$$${then}\:\:{x}^{\mathrm{3}} +\mathrm{3}{px}^{\mathrm{2}} +{qx}+{r}=\left({x}+{p}\right)^{\mathrm{3}} \\ $$

Commented by mr W last updated on 31/Jan/20

is this what you meant?

$${is}\:{this}\:{what}\:{you}\:{meant}? \\ $$

Commented by peter frank last updated on 31/Jan/20

i dont understand the question sir.what if the equation remain x^3 +3px+qx+r is it possible find relation r and q?

$${i}\:{dont}\:{understand}\:{the}\: \\ $$$${question}\:{sir}.{what}\:{if} \\ $$$${the}\:{equation}\:{remain} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{px}+{qx}+{r}\:{is}\:{it}\: \\ $$$${possible}\:{find}\:{relation} \\ $$$${r}\:{and}\:{q}? \\ $$

Commented by mr W last updated on 31/Jan/20

since q=3p^2 and r=p^3 you can say ((q/3))^3 =r^2 or q^3 =27r^2 .

$${since}\:{q}=\mathrm{3}{p}^{\mathrm{2}} \:{and}\:{r}={p}^{\mathrm{3}} \:{you}\:{can}\:{say} \\ $$$$\left(\frac{{q}}{\mathrm{3}}\right)^{\mathrm{3}} ={r}^{\mathrm{2}} \:{or}\:{q}^{\mathrm{3}} =\mathrm{27}{r}^{\mathrm{2}} . \\ $$

Commented by MJS last updated on 31/Jan/20

x^3 +3px^2 +qx+r=(x−z)^3 ⇒ z+p=0∧3z^2 −q=0∧z^3 +r=0 z=−p q=3p^2 r=p^3 ⇒ (q/r)=(3/p)

$${x}^{\mathrm{3}} +\mathrm{3}{px}^{\mathrm{2}} +{qx}+{r}=\left({x}−{z}\right)^{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$${z}+{p}=\mathrm{0}\wedge\mathrm{3}{z}^{\mathrm{2}} −{q}=\mathrm{0}\wedge{z}^{\mathrm{3}} +{r}=\mathrm{0} \\ $$$${z}=−{p} \\ $$$${q}=\mathrm{3}{p}^{\mathrm{2}} \\ $$$${r}={p}^{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$$\frac{{q}}{{r}}=\frac{\mathrm{3}}{{p}} \\ $$

Commented by peter frank last updated on 01/Feb/20

thank you @ mr w @MJS

$${thank}\:{you}\:@\:{mr}\:{w}\:@{MJS} \\ $$

Commented by mr W last updated on 01/Feb/20

MJS sir: i think (q/r)=(3/p) doesn′t ensure that x^3 +3px^2 +qx+r=(x−z)^3 . e.g. with p=2, q=±3, r=±2 we habe x^3 +3px^2 +qx+r=x^3 +6x^2 ±3x±2. but both ≠(x−z)^3 .

$${MJS}\:{sir}: \\ $$$${i}\:{think}\:\frac{{q}}{{r}}=\frac{\mathrm{3}}{{p}}\:{doesn}'{t}\:{ensure}\:{that} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{px}^{\mathrm{2}} +{qx}+{r}=\left({x}−{z}\right)^{\mathrm{3}} .\:{e}.{g}.\:{with} \\ $$$${p}=\mathrm{2},\:{q}=\pm\mathrm{3},\:{r}=\pm\mathrm{2}\:{we}\:{habe} \\ $$$${x}^{\mathrm{3}} +\mathrm{3}{px}^{\mathrm{2}} +{qx}+{r}={x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} \pm\mathrm{3}{x}\pm\mathrm{2}.\:{but} \\ $$$${both}\:\neq\left({x}−{z}\right)^{\mathrm{3}} . \\ $$

Commented by MJS last updated on 01/Feb/20

we need also q=3p^2 ∧r=p^3 p=2 ⇒ q=12∧r=8; z=−2 x^3 +6x^2 +12x+8=(x+2)^3

$$\mathrm{we}\:\mathrm{need}\:\mathrm{also}\:{q}=\mathrm{3}{p}^{\mathrm{2}} \wedge{r}={p}^{\mathrm{3}} \\ $$$${p}=\mathrm{2}\:\Rightarrow\:{q}=\mathrm{12}\wedge{r}=\mathrm{8};\:{z}=−\mathrm{2} \\ $$$${x}^{\mathrm{3}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{8}=\left({x}+\mathrm{2}\right)^{\mathrm{3}} \\ $$

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