A Very Nice Math Olympiad Problem | Solve for real value of x for which x^30+x^20=80 

this was an Olympiad prob???

Somebody loves 64 + 16 = 80. (x^10)^3 + (x^10)^2 = 80. Initially x^10 = 4, giving x = 2^(1/5) with another 9 roots to find.

2 problems 1. 4^(1/10) isn’t simplest form 4^(1/10) = (2^2)^(1/10) = 2^(2/10) = 2^(1/5) 2. There are 2 real solutions: x^10 = 4 x^10 - 4 = 0 (x^5)^2 - 2^2 = 0 Difference of squares (x^5 - 2)(x^5 + 2) = 0 x = +- 2^(1/5)

I would like to show the 2 points as follows: 1) There is another solution: x = - (4^(1/10)) [ There are 2 real solutions. ] 2) As for writing the final answer, 2^(1/5) is better than 4^(1/10) . Therefore, my final answer is x = ± 2^(1/5) . >

Or 10th root 4