Find-the-cube-root-of-26-5-3-

Question Number 52406 by Tawa1 last updated on 07/Jan/19

Find the cube root of 26 + 5 \sqrt{3}

Answered by MJS last updated on 07/Jan/19

no exact solution

I think it should be 26 + 15 \sqrt{3}

{(a + b \sqrt{3})}^{3} = 26 + 15 \sqrt{3}

(a^{3} + 9 {a b}^{2}) + 3 (a^{2} b + b^{3}) \sqrt{3} = 26 + 15 \sqrt{3}

a^{3} + 9 {a b}^{2} = 26 \Rightarrow b = \frac{\sqrt{26 - a^{3}}}{3 \sqrt{a}}

3 (a^{2} b + b^{3}) = 15 \Rightarrow

\Rightarrow \frac{2 (4 a^{3} + 13) \sqrt{26 - a^{3}}}{27 \sqrt{a^{3}}} = 5 \Rightarrow

\Rightarrow a^{9} - \frac{39}{2} a^{6} + \frac{8085}{64} a^{3} - \frac{2197}{8} = 0

a = \sqrt[3]{t}

t^{3} - \frac{39}{2} t^{2} - \frac{8085}{64} t - \frac{2197}{8} = 0

t = z + \frac{13}{2}

z^{3} - \frac{27}{64} z - \frac{351}{128} = 0

D = \frac{p^{3}}{27} + \frac{q^{2}}{4} > 0 \Rightarrow 1 real & 2 complex solutions

trying the factors of - \frac{351}{128} we find

z = \frac{3}{2} \Rightarrow t = 8 \Rightarrow a = 2 \Rightarrow b = 1

\sqrt[3]{26 + 15 \sqrt{3}} = 2 + \sqrt{3}

Commented by Tawa1 last updated on 07/Jan/19

God bless you sir .

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Jan/19

26 + 15 \sqrt{3}

= {(\sqrt{3})}^{3} + 12 \sqrt{3} + 26

= {(\sqrt{3})}^{3} + 3 . {(\sqrt{3})}^{2} . 2 + 3 . \sqrt{3} . {(2)}^{2} + {(2)}^{3}

= {(\sqrt{3} + 2)}^{3}

c u b e r o o t i s (2 + \sqrt{3})

Commented by Tawa1 last updated on 07/Jan/19

God bless you sir

Commented by malwaan last updated on 10/Jan/19

great

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