Solve-for-real-numbers-x-9-256x-3-791-84x-3-4x-7-1-3-

Question Number 154677 by mathdanisur last updated on 20/Sep/21

Solve for real numbers :

\frac{x^{9} - {256 x}^{3} - 791}{{84 x}^{3}} = \sqrt[3]{4 x + 7}

Commented by MJS_new last updated on 20/Sep/21

\frac{x^{9} - 256 x - 791}{84 x^{3}} = \sqrt[3]{4 x + 7}

we can solve it \dots

let x = \sqrt[3]{4 x + 7} \Rightarrow x^{3} - 4 x - 7 = 0 (1)

\Rightarrow

\frac{x^{9} - 256 x - 791}{84 x^{3}} = x

\Leftrightarrow

x^{9} - 84 x^{4} - 256 x - 791 = 0 (2)

(x^{3} - 4 x - 7) (x^{6} + 4 x^{4} + 7 x^{3} + 16 x^{2} - 28 x + 113) = 0

\Rightarrow

the solutions of (1) also solve (2)

x \in R \Rightarrow x = \sqrt[3]{\frac{7}{2} - \frac{\sqrt{3201}}{18}} + \sqrt[3]{\frac{7}{2} + \frac{\sqrt{3201}}{18}}

Commented by mathdanisur last updated on 20/Sep/21

There was a typo . Thank you for your attention S er .

The correct is \dots 256 x, not {256 x}^{3} .

Sorry for my mistake S er .

Commented by MJS_new last updated on 20/Sep/21

only approximation is possible

I get

x_{1} \approx - 2.05382774

x_{2} \approx - 1.30927779

x_{3} \approx 2.82876252

Commented by mathdanisur last updated on 20/Sep/21

creativ solution, thank you S er

Commented by mr W last updated on 20/Sep/21

{i t}^{'} s e a c h t i m e a w o n d e r t o m e h o w

y o u g e t

(x^{3} - 4 x - 7) (x^{6} + 4 x^{4} + 7 x^{3} + 16 x^{2} - 28 x + 113) = 0

f r o m

x^{9} - 84 x^{4} - 256 x - 791 = 0 .

g r e a t!

Commented by mathdanisur last updated on 20/Sep/21

Thank you so much S er

Commented by MJS_new last updated on 20/Sep/21

this time I first plotted both sides of the

given equation and it looked as if the only

solution was on the line y = x . then I tried to

verify this and it was easy to divide

(x^{9} - 84 x^{4} - 256 x - 791) / (x^{3} - 4 x - 7)

I see no chance to get the exact result

without this very lucky coincidence \dots

Commented by MJS_new last updated on 21/Sep/21

we can build similar questions this way :

\frac{x^{9} - a^{4} x - (a^{3} + b^{2}) b}{3 {a b x}^{3}} = \sqrt[3]{a x + b}

{it}^{'} s more a magic trick than serious math

let \sqrt[3]{a x + b} = x \Leftrightarrow x^{3} - a x - b = 0 (1)

\Rightarrow

\frac{x^{9} - a^{4} x - (a^{3} + b^{2}) b}{3 {a b x}^{3}} = x

x^{9} - 4 {a b x}^{4} - a^{4} x - (a^{3} + b^{2}) b = 0 (2)

(x^{3} - a x - b) (x^{6} + {a x}^{4} + {b x}^{3} + a^{2} x^{2} - a b x + a^{3} + b^{2}) = 0

obviously the real solution (s) of (1) also

solve (2) .

Commented by MJS_new last updated on 21/Sep/21

question 154303 works exactly the same way

Commented by mathdanisur last updated on 21/Sep/21

Very nice S er, thank you

Answered by imjagoll last updated on 21/Sep/21

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