find-the-cubic-of-26-15-3-1-3-26-15-3-1-3-

Question Number 172022 by Mikenice last updated on 23/Jun/22

f i n d t h e c u b i c o f

\sqrt[3]{26 + 15 \sqrt{3}} + \sqrt[3]{26 - 15 \sqrt{3}}

Answered by Rasheed.Sindhi last updated on 23/Jun/22

L e t \sqrt[3]{26 + 15 \sqrt{3}} + \sqrt[3]{26 - 15 \sqrt{3}} = a

w h e r e a \in R

{(\sqrt[3]{26 + 15 \sqrt{3}} + \sqrt[3]{26 - 15 \sqrt{3}})}^{3} = a^{3}

\begin{matrix} {(a + b)}^{3} = a^{3} + b^{3} + 3 a b (a + b) \end{matrix}

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

(26 + 15 \sqrt{3}) (26 - 15 \sqrt{3})

= 26^{2} - 3 (15^{2}) = 676 - 675 = 1

52 + 3 (1) (a) = a^{3}

a^{3} - 3 a - 52 = 0

(a - 4) (a^{2} + 4 a + 13) = 0

a = 4 ∣ a = \frac{- 4 \pm \sqrt{16 - 52}}{2} = \frac{- 4 \pm 6 i}{2} = - 2 \pm 3 i \notin R (R e j e c t e d)

a = 4 (o n l y s o l u t i o n)

Commented by Mikenice last updated on 23/Jun/22

t h a n k s s i r

Commented by mr W last updated on 23/Jun/22

r e a l + r e a l = r e a l

\sqrt[3]{26 + 15 \sqrt{3}} + \sqrt[3]{26 - 15 \sqrt{3}} = r e a l

\Rightarrow a n s w e r c a n o n l y b e r e a l!

Commented by Rasheed.Sindhi last updated on 23/Jun/22

T han k s sir! I^{'} m going to correct now .

Commented by Mikenice last updated on 23/Jun/22

b u t s i r [y o u c a n s h o w y o u r o w n

s o l u t i o n t o t h a t p r o b l e m .

Commented by mr W last updated on 23/Jun/22

i {d o n}^{'} t h a v e a n o t h e r s o l u t i o n . t h e

s o l u t i o n f r o m r a s h e e d i s v e r y g o o d .

Commented by Mikenice last updated on 23/Jun/22

o k a y s i r