Simplify: _3 (√(3+((10)/3)(√(1/3))i))+ _3 (√(3−((10)/3)(√(1/3))i))


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Question Number 78800 by Tony Lin last updated on 20/Jan/20

$S i m p l i f y :$ $\underset{3}{} \sqrt{3 + \frac{10}{3} \sqrt{\frac{1}{3}} i} + \underset{3}{} \sqrt{3 - \frac{10}{3} \sqrt{\frac{1}{3}} i}$
Commented by mathmax by abdo last updated on 20/Jan/20

$A = {(3 + \frac{10}{3 \sqrt{3}} i)}^{\frac{1}{3}} + {(3 - \frac{10}{3 \sqrt{3}} i)}^{\frac{1}{3}}$ $w e h a v e ∣ 3 + \frac{10}{3 \sqrt{3}} i ∣ = \sqrt{3^{2} + \frac{10^{2}}{27}} = \sqrt{\frac{37 \times 9 + 100}{27}} = \sqrt{\frac{433}{37} = r_{0}}$ $\Rightarrow 3 + \frac{10}{3 \sqrt{3}} i = r_{0} e^{i a r c t a n (\frac{10}{9 \sqrt{3}})} \Rightarrow {(3 + \frac{10}{3 \sqrt{3}} i)}^{\frac{1}{3}} = {(r_{0})}^{\frac{1}{3}} e^{\frac{i}{3} a r c t a n (\frac{10}{9 \sqrt{3}})}$ $a l s o {(3 - \frac{10}{3 \sqrt{3}} i)}^{\frac{1}{3}} = {(r_{0})}^{\frac{1}{3}} e^{- \frac{i}{3} a r c t a n (\frac{10}{9 \sqrt{3}})} \Rightarrow$ $A = {(r_{0})}^{\frac{1}{3}} (2 c o s (\frac{1}{3} a r c t a n (\frac{10}{9 \sqrt{3}})) = 2 {(r_{0})}^{\frac{1}{3}} c o s (\frac{1}{3} a r c t a n (\frac{10}{9 \sqrt{3}}))$
Commented by Tony Lin last updated on 21/Jan/20

$t h a n k s s i r, b u t r_{0} = \sqrt{\frac{27 \times 9 + 100}{27}} = \sqrt{\frac{343}{27}}$
Commented by mathmax by abdo last updated on 21/Jan/20

$y e s$
	Answered by MJS last updated on 20/Jan/20

	$x = a^{\frac{1}{3}} + b^{\frac{1}{3}} ∣^{3}$ $x^{3} = a + 3 a^{\frac{1}{3}} b^{\frac{1}{3}} (a^{\frac{1}{3}} + b^{\frac{1}{3}}) + b$ $x^{3} = a + 3 a^{\frac{1}{3}} b^{\frac{1}{3}} x + b$ $x^{3} - 3 a^{\frac{1}{3}} b^{\frac{1}{3}} x - (a + b) = 0$ $a = u + v \land b = u - v$ $x^{3} - 3 {(u^{2} - v^{2})}^{\frac{1}{3}} x - 2 u = 0$ $u = 3 \land v = \frac{10}{3} \sqrt{\frac{1}{3}} i$ $x^{3} - 7 x - 6 = 0$ $\Rightarrow$ $x_{1} = - 2$ $x_{2} = - 1$ $x_{3} = 3$ $the angles of 3 \pm \frac{10}{3} \sqrt{\frac{1}{3}} i are$ $\pm \arctan \frac{10 \sqrt{3}}{27} \approx \pm . 570377; their absolute$ $value is smaller than \frac{π}{2} \Rightarrow the solution must$ $be > 0 \Rightarrow solution is$ $\sqrt[3]{3 + \frac{10}{3} \sqrt{\frac{1}{3}} i} + \sqrt[3]{3 - \frac{10}{3} \sqrt{\frac{1}{3}} i} = 3$
	Commented by Tony Lin last updated on 21/Jan/20

	$t h a n k s s i r, b u t i f w e o r i g i n a l l y u s e$ ${C a r d a n o}^{'} s m e t h o d t o s o l v e$ $x^{3} + p x + q = 0$ $h o w c a n w e s i m p l i f y t h e a n s w e r t o$ $b e c o m e m o r e b e a u t i f u l w i t h o u t u s i n g$ $f a c t o r i z a t i o n$ $o r w h a t w e c a n d o i s o n l y c u b e i t$ $a n d s o l v e i t i n a c o m p l i c a t e d a n d$ $s l o w w a y$
	Commented by MJS last updated on 21/Jan/20

	$Cardano {doesn}^{'} t work when the equation has$ $3 real roots .$ $x^{3} + p x + q = 0$ $first, always check all factors of \pm q$ $(x - α) (x - β) (x - γ) = x^{3} + . . . - α β γ$ $\Rightarrow q = - α β γ$ $if we have solutions \in Z or \in Q we can simply$ $find them$ $second, calculate D = \frac{p^{3}}{27} + \frac{q^{2}}{4}$ $D ⩾ 0 \Rightarrow Cardano$ $D < 0 \Rightarrow trigonometric solution$ $x_{k + 1} = \frac{2}{3} \sqrt{- 3 p} \sin (\frac{1}{3} (2 k π + \arcsin \frac{9 q}{2 \sqrt{- 3 p^{3}}}))$ $with k = 0, 1, 2$ $the problem with both methods is that the$ $solutions ‘ ‘ {hide}^{″} easier ones like x = \sqrt{3} + \sqrt{2}$ $but {there}^{'} s no other way$
	Commented by MJS last updated on 21/Jan/20
	$x^3 −7x−6=0 in this case, first try factors of ±6: {−6, −3, −2, −1, 1, 2, 3, 6} and you′re done Cardano doesn′t work because D=(((−7)^3 )/(27))+(((−6)^2 )/4)=−((100)/(27))<0 trigonometric solution gives x_1 =−((√(21))/3)sin ((1/3)arcsin ((9(√(21)))/(49))) x_2 =((2(√(21)))/3)sin ((π/3)+arcsin ((9(√(21)))/(49))) x_3 =−((2(√(21)))/3)cos ((π/6)+arcsin ((9(√(21)))/(49))) and it′s not easy to show x_1 =−1 x_2 =3 x_3 =−2$
	$x^{3} - 7 x - 6 = 0$ $in this case, first try factors of \pm 6 :$ ${- 6, - 3, - 2, - 1, 1, 2, 3, 6}$ $and {you}^{'} re done$ $Cardano {doesn}^{'} t work because$ $D = \frac{{(- 7)}^{3}}{27} + \frac{{(- 6)}^{2}}{4} = - \frac{100}{27} < 0$ $trigonometric solution gives$ $x_{1} = - \frac{\sqrt{21}}{3} \sin (\frac{1}{3} \arcsin \frac{9 \sqrt{21}}{49})$ $x_{2} = \frac{2 \sqrt{21}}{3} \sin (\frac{π}{3} + \arcsin \frac{9 \sqrt{21}}{49})$ $x_{3} = - \frac{2 \sqrt{21}}{3} \cos (\frac{π}{6} + \arcsin \frac{9 \sqrt{21}}{49})$ $and {it}^{'} s not easy to show$ $x_{1} = - 1$ $x_{2} = 3$ $x_{3} = - 2$
	Commented by jagoll last updated on 21/Jan/20

	$waw . . i understand sir . thanks sir$
	Commented by Tony Lin last updated on 21/Jan/20

	$t h a n k s s i r, a n d I f o u n d t h a t$ $i f I u s e t h e r e s u l t f r o m m a t h m a x$ $\sqrt[3]{3 + \frac{10}{3} \sqrt{\frac{1}{3}} i} + \sqrt[3]{3 - \frac{10}{3} \sqrt{\frac{1}{3}} i}$ $= 2 \sqrt{\frac{7}{3}} c o s [\frac{1}{3} a r c t a n (\frac{10}{9 \sqrt{3}})]$ $l e t a r c t a n (\frac{10}{9 \sqrt{3}}) = θ \Rightarrow t a n θ = \frac{10}{9 \sqrt{3}}$ $∴ c o s θ = \frac{9 \sqrt{3}}{7 \sqrt{7}} = 4 {c o s}^{3} \frac{θ}{3} - 3 c o s \frac{θ}{3}$ $l e t c o s \frac{θ}{3} = t \Rightarrow 4 t^{3} - 3 t = \frac{9 \sqrt{3}}{7 \sqrt{7}}$ $\Rightarrow 28 \sqrt{7} t^{3} - 21 \sqrt{7} t - 9 \sqrt{3} = 0$ $\Rightarrow (2 \sqrt{7} t - 3 \sqrt{3}) (14 t^{2} + 3 \sqrt{21} t + 3) = 0$ $∴ t = \frac{3}{2} \sqrt{\frac{3}{7}}$ $∴ 2 \sqrt{\frac{7}{3}} c o s [\frac{1}{3} a r c t a n (\frac{10}{9 \sqrt{3}})]$ $= \sqrt[3]{3 + \frac{10}{3} \sqrt{\frac{1}{3}} i} + \sqrt[3]{3 - \frac{10}{3} \sqrt{\frac{1}{3}} i}$ $= 3$
	Commented by msup trace by abdo last updated on 21/Jan/20

	$t h a n k y o u f o r c o m p l e t i n g t h e a n s w e r$
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