Russia Math Olympiad Problem With Cubic Radicals | Simplifying Cube Root Radicals Without Cal.

Russia Math Olympiad Problem With Cubic Radicals.
In this video I will show you how to simplify radicals with cube roots without the use of the calculator in few steps.
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[@michaeldoerr5810]

That is an exmaple of easuer than it looks. Thia reminds me of cubic equarions where two-thirds of the roots are complex and one-third are real. If you are looking for a real soln it is just 1 real solution. And that real soln is x = 2. The complex solutions are -1+/-sqrt(6)i.

[@RyanLewis-Johnson-wq6xs]

It’s in my head.

[@satrajitghosh8162]

writing a ^3 =7 + √(50) b ^ 3 = ( 7 - √(50)) one gets a * b = 49 - 50 = -1 Hereby a ^3 + b ^ 3 = 14 ( a + b) ^3 - 3 a b ( a + b) = 14 ( a + b)^3 + 3 ( a + b) = 14 Writing z = a + b one gets z^3 - 2 ^ 3 + 3 ( z - 2) = 0 (z - 2)( z^2 + 2 z + 4 + 3) = 0 a + b = z = 2, - 1 + i √6, - 1 - i √6,

[@danielfranca1939]

The move in this solving is lively man. Keep it up...❤

[@yakupbuyankara5903]

2

[@letstalkwithbigsheila]

Getting more interested in maths

[@danielfranca1939]

Same here 😂😅😂

[@RyanLewis-Johnson-wq6xs]

Cbrt[7+Sqrt[50]]+Cbrt[7-Sqrt[50]]=2 real root final answer 1±Sqrt[6]i