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

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

F i n d t h e r e l a t i o n b e t w e e n

q a n d r s o t h a t

x^{3} + 3 {p x}^{2} + q x + r i s a p e r f e c t

c u b e f o r a l l v a l u e o f x

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

y e s s i r

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

y o u m e a n x^{3} + 3 {p x}^{2} + q x + r ?

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

\Rightarrow q = 3 p^{2}

\Rightarrow r = p^{3}

t h e n x^{3} + 3 {p x}^{2} + q x + r = {(x + p)}^{3}

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

i s t h i s w h a t y o u m e a n t ?

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

i d o n t u n d e r s t a n d t h e

q u e s t i o n s i r . w h a t i f

t h e e q u a t i o n r e m a i n

x^{3} + 3 p x + q x + r i s i t

p o s s i b l e f i n d r e l a t i o n

r a n d q ?

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

s i n c e q = 3 p^{2} a n d r = p^{3} y o u c a n s a y

{(\frac{q}{3})}^{3} = r^{2} o r q^{3} = 27 r^{2} .

Commented by MJS last updated on 31/Jan/20

x^{3} + 3 {p x}^{2} + q x + r = {(x - z)}^{3}

\Rightarrow

z + p = 0 \land 3 z^{2} - q = 0 \land z^{3} + r = 0

z = - p

q = 3 p^{2}

r = p^{3}

\Rightarrow

\frac{q}{r} = \frac{3}{p}

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

t h a n k y o u @ m r w @ M J S

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

M J S s i r :

i t h i n k \frac{q}{r} = \frac{3}{p} {d o e s n}^{'} t e n s u r e t h a t

x^{3} + 3 {p x}^{2} + q x + r = {(x - z)}^{3} . e . g . w i t h

p = 2, q = \pm 3, r = \pm 2 w e h a b e

x^{3} + 3 {p x}^{2} + q x + r = x^{3} + 6 x^{2} \pm 3 x \pm 2 . b u t

b o t h \neq {(x - z)}^{3} .

Commented by MJS last updated on 01/Feb/20

we need also q = 3 p^{2} \land r = p^{3}

p = 2 \Rightarrow q = 12 \land r = 8; z = - 2

x^{3} + 6 x^{2} + 12 x + 8 = {(x + 2)}^{3}

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