Simplify-3-3-10-3-1-3-i-3-3-10-3-1-3-i-

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} + \dots - 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} - 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