If , (( 1 − lo^ g_( 2) (x)))^(1/3) + ((1+^ log_( 2) (x)))^(1/3) −1=0 ⇒ x = ?


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Question Number 186419 by mnjuly1970 last updated on 04/Feb/23

$I f, \sqrt[3]{1 - l {\overset{}{o g}}_{2} (x)} + \sqrt[3]{1 \overset{}{+} {l o g}_{2} (x)} - 1 = 0$ $\Rightarrow x = ?$
	Answered by cortano1 last updated on 04/Feb/23
	$We know that if a+b+c = 0 then a^3 +b^3 +c^3 = 3abc ⇒((1−log _2 x))^(1/3) +((1+log _2 x))^(1/3) +(−1)=0 ⇒(1−log _2 x)+(1+log _2 x)+(−1)=−3(((1−(log _2 x)^2 ))^(1/3) ⇒−(1/3) = ((1−(log _2 x)^2 ))^(1/3) ⇒(log _2 x)^2 = ((28)/(27)) ⇒ { ((x=2^(√((28)/(27))) =2^((2(√7))/(3(√3))) )),((x= (1/((2^((2(√7))/( 3(√3))) ))))) :}$
	$W e k n o w t h a t i f a + b + c = 0$ $t h e n a^{3} + b^{3} + c^{3} = 3 a b c$ $\Rightarrow \sqrt[3]{1 - \log_{2} x} + \sqrt[3]{1 + \log_{2} x} + (- 1) = 0$ $\Rightarrow (1 - \log_{2} x) + (1 + \log_{2} x) + (- 1) = - 3 \sqrt[3]{(1 - {(\log_{2} x)}^{2}}$ $\Rightarrow - \frac{1}{3} = \sqrt[3]{1 - {(\log_{2} x)}^{2}}$ $\Rightarrow {(\log_{2} x)}^{2} = \frac{28}{27}$ $\Rightarrow {\begin{cases} x = 2^{\sqrt{\frac{28}{27}}} = 2^{\frac{2 \sqrt{7}}{3 \sqrt{3}}} \\ x = \frac{1}{(2^{\frac{2 \sqrt{7}}{3 \sqrt{3}}})} \end{cases}$
	Commented by mnjuly1970 last updated on 04/Feb/23

	$t h a n k s a l o t s i r c o r t s n o$
	Answered by mahdipoor last updated on 04/Feb/23

	$x = 2^{y^{3} - 1} \Rightarrow^{3} \sqrt{2 - y^{3}} + y - 1 = 0 \Rightarrow$ ${(^{3} \sqrt{2 - y^{3}})}^{3} = {(1 - y)}^{3} \Rightarrow 2 - y^{3} = 1 - 3 y + 3 y^{2} - y^{3}$ $\Rightarrow 3 y^{2} - 3 y - 1 = 0 \Rightarrow y = \frac{3 \pm \sqrt{21}}{6} = \frac{1}{2} \pm \frac{\sqrt{21}}{6}$ $y^{3} - 1 = \frac{1}{8} \pm \frac{\sqrt{21}}{8} + \frac{21}{24} \pm \frac{7 \sqrt{21}}{72} - 1 = \pm \frac{2}{9} \sqrt{21}$ $\Rightarrow x = 2^{\land} (\pm \frac{2 \sqrt{21}}{9})$
	Commented by mnjuly1970 last updated on 04/Feb/23

	$m a n o o n o s t a d a l i b o o d$
	Answered by Rasheed.Sindhi last updated on 04/Feb/23

	$\sqrt[3]{1 - \log_{2} x} + \sqrt[3]{1 + \log_{2} x} - 1 = 0$ ${(\sqrt[3]{1 - \log_{2} x} + \sqrt[3]{1 + \log_{2} x})}^{3} = 1$ $1 - \log_{2} x + 1 + \log_{2} x + 3 (1 - \log_{2} x) (1 + \log_{2} x) (\sqrt[3]{1 - \log_{2} x} + \sqrt[3]{1 + \log_{2} x}) = 1$ $2 + 3 (\sqrt[3]{1 - \log_{2} x}) (\sqrt[3]{1 + \log_{2} x}) (1) = 1$ $2 + 3 (\sqrt[3]{1 - {(\log_{2} x)}^{2}}) = 1$ $\sqrt[3]{1 - {(\log_{2} x)}^{2}} = - 1 / 3$ $1 - {(\log_{2} x)}^{2} = - \frac{1}{27}$ ${(\log_{2} x)}^{2} = 1 + \frac{1}{27} = \frac{28}{27}$ $\log_{2} x = \frac{\pm 2 \sqrt{7}}{3 \sqrt{3}} \cdot \frac{\pm 2 \sqrt{21}}{9}$ $x = 2^{\frac{\pm 2 \sqrt{21}}{9}}$
	Commented by mnjuly1970 last updated on 04/Feb/23

	$t h a n k y o u s o m u c h$ $s i r r a s h e e d$
	Commented by mahdipoor last updated on 04/Feb/23

	${(a + b)}^{3} = a^{3} + b^{3} + 3 a b (a + b) =$ $. . . + 3^{3} \sqrt{1 - {l o g}_{2} x} \times^{3} \sqrt{1 + {l o g}_{2} x} \times (. . .)$
	Commented by Rasheed.Sindhi last updated on 04/Feb/23

	$T h a n k y o u s i r, I^{'} v e c o r r e c t e d .$
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