Correct answer: Two irrational and one rational solutions
Let a=2+t, b=t-1, c=1+2t=a+b. Equation becomes a³+b³=c³=(a+b)³. So 3ab(a+b)=0 => ab=0 or a+b=0. ab=0: (2+t)(t-1)=0 => t=-2 or t=1 => x=0.01 or x=10 (both rational). a+b=0: 2t+1=0 => t=-1/2 => x=10^(-1/2)=1/sqrt(10) (irrational). So three solutions: two rational (x=1/100 and x=10) and one irrational (x=1/sqrt(10)). Wait — 'two rational and one irrational' matches option B. But option A says 'two irrational and one rational'. Recheck: t=-2 -> x=10^(-2)=0.01 (rational), t=1 -> x=10 (rational), t=-1/2 -> x=10^(-1/2) (irrational). So: 1 irrational + 2 rational. Answer is option B.