3

Question

3

Answer 1

(x+1)^(1/6) = t

t^2-3t+2=0

D=1

t1=2   t2=1

(x+1)^(1/6)  = 1

x+1 = 1    x1=0

(x+1)^(1/6)  = 2

x+1 = 2^6

x+1=64

x2=63