BeMath-x-4-1-x-4-23-x-3-1-x-3-

Question Number 107945 by bemath last updated on 13/Aug/20

\frac{\circ B e M a t h \circ}{\land ⌣ \land}

{\begin{cases} x^{4} + \frac{1}{x^{4}} = 23 \\ x^{3} - \frac{1}{x^{3}} = ? \end{cases}

Answered by $@y@m last updated on 13/Aug/20

{(x^{2} + \frac{1}{x^{2}})}^{2} - 2 = 23

{(x^{2} + \frac{1}{x^{2}})}^{2} = 25

x^{2} + \frac{1}{x^{2}} = 5

x^{2} + \frac{1}{x^{2}} - 2 = 3

{(x - \frac{1}{x})}^{2} = 3

x - \frac{1}{x} = \sqrt{3}

{(x - \frac{1}{x})}^{3} = 3 \sqrt{3}

x^{3} - \frac{1}{x^{3}} - 3 (x - \frac{1}{x}) = 3 \sqrt{3}

x^{3} - \frac{1}{x^{3}} - 3 \sqrt{3} = 3 \sqrt{3}

x^{3} - \frac{1}{x^{3}} = 6 \sqrt{3}

Commented by Rasheed.Sindhi last updated on 13/Aug/20

Nic e!

Commented by Her_Majesty last updated on 13/Aug/20

n i c e b u t a l s o - 6 \sqrt{3} i s a s o l u t i o n

Commented by bemath last updated on 13/Aug/20

y e s s i r . t h e a n s w e r \pm 6 \sqrt{3}

Answered by 1549442205PVT last updated on 13/Aug/20

x^{4} + \frac{1}{x^{4}} = 23 \Leftrightarrow {(x^{2} + \frac{1}{x^{2}})}^{2} = 25

\Leftrightarrow x^{2} + \frac{1}{x^{2}} = \pm 5

i) x^{2} + \frac{1}{x^{2}} = 5 \Leftrightarrow {(x - \frac{1}{x})}^{2} = 3 \Leftrightarrow x - \frac{1}{x} = \pm \sqrt{3}

a) x - \frac{1}{x} = \sqrt{3} \Rightarrow x^{3} - \frac{1}{x^{3}} = {(x - \frac{1}{x})}^{3} + 3 (x - \frac{1}{x})

= 3 \sqrt{3} + 3 \sqrt{3} = 6 \sqrt{3}

b) x - \frac{1}{x} = - \sqrt{3} \Rightarrow x^{3} - \frac{1}{x^{3}} = {(- \sqrt{3})}^{3} + 3 . (- \sqrt{3})

= - 6 \sqrt{3}

ii) x^{2} + \frac{1}{x^{2}} = - 5 \Rightarrow {(x - \frac{1}{x})}^{2} = - 7

\Rightarrow x - \frac{1}{x} = \pm i \sqrt{7}

c) x - \frac{1}{x} = i \sqrt{7} \Rightarrow x^{3} - \frac{1}{x^{3}} = {(i \sqrt{7})}^{3} + 3 i \sqrt{7}

= - 7 i \sqrt{7} + 3 i \sqrt{7} = - 4 i \sqrt{7}

d) x - \frac{1}{x} = - i \sqrt{7} \Rightarrow x^{3} - \frac{1}{x^{3}} = {(- i \sqrt{7})}^{3} - 3 i \sqrt{7}

= 7 i \sqrt{7} - 3 i \sqrt{7} = 4 i \sqrt{7}

Thus, we get four answers follows as :

x^{3} - \frac{1}{x^{3}} \in {6 \sqrt{3}; - 6 \sqrt{3}; 4 i \sqrt{7}; - 4 i \sqrt{7}}

Answered by Rasheed.Sindhi last updated on 13/Aug/20

An uncommon {way}_{}^{} \to

{\begin{cases} x^{4} + \frac{1}{x^{4}} = 23 \\ x^{3} - \frac{1}{x^{3}} = ? \end{cases}

{(x^{4} + \frac{1}{x^{4}})}^{3} = {(23)}^{3}

x^{12} + \frac{1}{x^{12}} + 3 (x^{4} + \frac{1}{x^{4}}) = 23^{3}

x^{12} + \frac{1}{x^{12}} = 23^{3} - 3 (23) = 12098

{(x^{6} + \frac{1}{x^{6}})}^{2} - 2 = 12098

{(x^{6} + \frac{1}{x^{6}})}^{2} = 12100

x^{6} + \frac{1}{x^{6}} = \pm 110

x^{6} + \frac{1}{x^{6}} - 2 = \pm 110 - 2

{(x^{3} - \frac{1}{x^{3}})}^{2} = 108, 112

x^{3} - \frac{1}{x^{3}} = \pm 6 \sqrt{3}, \pm 4 \sqrt{7}

\underset{↓}{\overset{↑}{⟨ ⋊_{}^{} {sindhi}_{}^{} ⋉ ⟩}}_{}^{}

Commented by bemath last updated on 13/Aug/20

c o o l l

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