Question

simplify :
∛(22√5 + 15√11) + ∛(22√5 - 15√11)

Answer 1

give me a clue for do this??

Answer 2

honestly..you cant do anything about this...

Answer 3

i was also thinking the same..

Answer 4

use \[(a+b)^{3} = a ^{3}+b ^{3}+3\times a \times b(a+b)\]
Put a+b = t where a+b is ur question.
solve the cubic.

Answer 5

sorri, the problem should be ∛(22√5 + 15√11) + ∛(22√5 - 15√11) - \[\sqrt[6]{3520}\], i want do it step by step only, hehe ... can u exlaind to me?

Answer 6

\[\huge \sqrt[?]{3520} ?\]

Answer 7

? is 6

Answer 8

let a=22√5 + 15√11 and b=22√5 - 15√11
ab=-55=-5*11

Answer 9

there is something wrong with equation editor
let me reload the page

Answer 10

\[\Large 22\sqrt{5} = 2\sqrt{11}\sqrt{55}\]
\[\Large15\sqrt{11} = 3\sqrt{5}\sqrt{55}\]
\[\Large \sqrt[6]{3520} = 2\sqrt[6]{55} = 2\sqrt[3]{\sqrt{55}}\]

\[\Large \sqrt[3]{(22\sqrt{5}+15\sqrt{11})}+\sqrt[3]{(22\sqrt{5}-15\sqrt{11})}-\sqrt[6]{3520}\]
\[\Large \sqrt[3]{(2\sqrt{11}\sqrt{55}+3\sqrt{5}\sqrt{55})}+\sqrt[3]{(2\sqrt{11}\sqrt{55}-3\sqrt{5}\sqrt{55})}-2\sqrt[3]{\sqrt{55}}\]
\[\Large A = \sqrt{55}\]
\[\Large \sqrt[3]{(2\sqrt{11}A+3\sqrt{5}A)}+\sqrt[3]{(2\sqrt{11}A-3\sqrt{5}A)}-2\sqrt[3]{A}\]
\[\Large \sqrt[3]{A(2\sqrt{11}+3\sqrt{5})}+\sqrt[3]{A(2\sqrt{11}-3\sqrt{5})}-2\sqrt[3]{A}\]
\[\Large \sqrt[3]{A}\sqrt[3]{(2\sqrt{11}+3\sqrt{5})}+\sqrt[3]{A}\sqrt[3]{(2\sqrt{11}-3\sqrt{5})}-2\sqrt[3]{A}\]
\[\Large \sqrt[3]{A}(\sqrt[3]{(2\sqrt{11}+3\sqrt{5})}+\sqrt[3]{(2\sqrt{11}-3\sqrt{5})}-2)\]

Another way to look at this:
\[\Large B = 2\sqrt{11} \]
\[\Large C = 3\sqrt{5}\]

\[\Large \sqrt[3]{A}(\sqrt[3]{(B+C)}+\sqrt[3]{(B-C)}-2)\]

That's about as far as I got. Can't recall if the sum and difference of cubes equations can be used on cube roots.