Find-the-cube-root-of-26-15-3-

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 \dots

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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