Find the cube root of 26 − 15(√3)


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Question Number 41522 by Tawa1 last updated on 09/Aug/18

$Find the cube root of 26 - 15 \sqrt{3}$
Commented by math khazana by abdo last updated on 10/Aug/18

$i n s i d e C z^{3} = 26 - 15 \sqrt{3} l e t z = x + i y \Rightarrow$ $x^{3} + 3 x^{2} i y + 3 x {(i y)}^{2} - {i y}^{3} = 26 - 15 \sqrt{3} \Rightarrow$ $x^{3} - 3 {x y}^{2} + i (3 x^{2} y - y^{3}) = 26 - 15 \sqrt{3} \Rightarrow$ $x^{3} - 3 {x y}^{2} = 26 - 15 \sqrt{3} a n d 3 x^{2} y - y^{3} = 0 \Rightarrow 3 x^{2} = y^{2}$ $\Rightarrow x^{3} - 9 x^{3} = 26 - 15 \sqrt{3} \Rightarrow - 8 x^{3} = 26 - 15 \sqrt{3}$ $l e t f i n d a a n d b / {(a - b \sqrt{3})}^{3} = 26 - 15 \sqrt{3} a f t e r$ $d e v e l o p p e m e n t w e g e t a = 2 a n d b = 1 \Rightarrow$ ${(2 - \sqrt{3})}^{3} = 26 - 15 \sqrt{3} \Rightarrow - 8 x^{3} = {(2 - \sqrt{3})}^{3} \Rightarrow$ $x^{3} = {(\frac{\sqrt{3} - 2}{2})}^{3} \Rightarrow x = \frac{\sqrt{3}}{2} - 1 a n d y = \bar{+} \sqrt{3} x$ $= \bar{+} \sqrt{3} (\frac{\sqrt{3}}{2} - 1) \Rightarrow$ $z = \frac{\sqrt{3}}{2} - 1 + i ξ (\frac{3}{2} - \sqrt{3}) w i t h ξ^{2} = 1$
Commented by Tawa1 last updated on 10/Aug/18

$God bless you sir$
	Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18

	$a^{3} - 3 a^{2} b + 3 {a b}^{2} - b^{3}$ $2^{3} - 3 . 2^{2} . \sqrt{3} + 3.2 . {(\sqrt{3})}^{2} - {(\sqrt{3})}^{3}$ $= {(2 - \sqrt{3})}^{3}$ $s o c u b e r o o t i s (2 - \sqrt{3})$
	Commented by Joel578 last updated on 09/Aug/18

	$How did u know a = 2 and b = \sqrt{3} ?$
	Commented by $@ty@m last updated on 09/Aug/18

	$- 3 a^{2} b - b^{3} = - 15 \sqrt{3}$ $i t i s p o s s i b l e o n l y i f b = \sqrt{3}$ $\Rightarrow 3 a^{2} \sqrt{3} + 3 \sqrt{3} = 15 \sqrt{3}$ $\Rightarrow 3 a^{2} \sqrt{3} = 12 \sqrt{3}$ $\Rightarrow a = 2$
	Commented by Joel578 last updated on 09/Aug/18

	$u n d e r s t o o d . t h a n k y o u v e r y m u c h$
	Answered by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18

	$o r m e t h o d$ $26 - 15 \sqrt{3} = {(a - b \sqrt{3})}^{3}$ $a^{3} - 3 a^{2} b \sqrt{3} + 3 a . b^{2} . 3 - 3 \sqrt{3} b^{3} = 26 - 15 \sqrt{3}$ $(a^{3} + 9 {a b}^{2}) - \sqrt{3} (3 a^{2} b + 3 b^{3}) = 26 - 15 \sqrt{3}$ $s o a^{3} + 9 {a b}^{2} = 26$ $3 a^{2} b + 3 b^{3} = 15$ $n o w s o l v e t o f i n d a a n d b$ $b (a^{2} + b^{2}) = 5$ $a (a^{2} + 9 b^{2}) = 26$ $\frac{a (a^{2} + 9 b^{2}}{b (a^{2} + b^{2})} = \frac{26}{5}$ $\frac{a}{b} \times \frac{(\frac{a^{2}}{b^{2}}) + 9}{(\frac{a^{2}}{b^{2}}) + 1} = \frac{26}{5}$ $l e t k = \frac{a}{b}$ $k . \frac{k^{2} + 9}{k^{2} + 1} = \frac{26}{5}$ $5 (k^{3} + 9 k) = 26 k^{2} + 26$ $5 k^{3} - 26 k^{2} + 45 k - 26 = 0$ $5 k^{3} - 10 k^{2} - 16 k^{2} + 32 k + 13 k - 26 = 0$ $5 k^{2} (k - 2) - 16 k (k - 2) + 13 (k - 2) = 0$ $(k - 2) (5 k^{2} - 16 k + 13) = 0$ $s o k = 2 t h a t m e a n s \frac{a}{b} = 2$ $b (a^{2} + b^{2}) = 5$ $b (4 b^{2} + b^{2}) = 5$ $5 b^{3} = 1$ $s o b = 1$ $a = 2 b t h a t m e a n s a = 2 \times 1 = 2$ $26 - 15 \sqrt{3} = {(a - b \sqrt{3})}^{3}$ $26 - 15 \sqrt{3} = {(2 - \sqrt{3})}^{3}$
	Commented by Tawa1 last updated on 09/Aug/18

	$Wow, God bless you sir .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 09/Aug/18

	$t h a n k u . . .$
	Commented by Joel578 last updated on 09/Aug/18

	$Just a little mistake Sir$ $5 b^{3} = 5 \to b^{3} = 1 \to b = 1$
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