解方程三次根号x+45 +三次根号16-x=1

设a=(x+45)^(1/3),b=(16-x)^(1/3)
则有a+b=1,a^3+b^3=45+16=61
将b=1-a带入后式得
a^3+(1-a)^3=61
a^3+(1-a)^3=a^3-a^3+3a^2-3a+1=3a^2-3a+1=61
a^2-a-20=0
得a=5,-4
则(x+45)^(1/3)=5,-4
解得x=80,-109