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

$$Findtherelationbetween$$$$qandrsothat$$$$x^3+3{px}^2+qx+risaperfect$$$$cubeforallvalueofx$$

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

$$yessir$$

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

$$youmean{x}^3+3{px}^2+qx+r?$$

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

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

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

$$isthiswhatyoumeant?$$

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

$$idontunderstandthe$$$$questionsir.whatif$$$$theequationremain$$$$x^3+3px+qx+risit$$$$possiblefindrelation$$$$randq?$$

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

$$sinceq=3p^{2}andr=p^{3}youcansay$$$${\left(\frac{q}{3}\right)}^3=r^{2}or{q}^3=27r^2.$$

Commented by MJS last updated on 31/Jan/20

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

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

$$thankyou@mrw@MJS$$

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

$$MJSsir:$$$$ithink\frac{q}{r}=\frac{3}{p}{doesn}^{\prime }tensurethat$$$$x^3+3{px}^2+qx+r={(x-z)}^3.e.g.with$$$$p=2,q=\pm 3,r=\pm 2wehabe$$$$x^3+3{px}^2+qx+r=x^3+6x^2\pm 3x\pm 2.but$$$$both\ne {(x-z)}^3.$$

Commented by MJS last updated on 01/Feb/20

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

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