设实数p=三次根号4－三次根号6＋三次根号9，求证1<p<2

p=2^(2/3)-2^(1/3)*3^(1/3)+3^(2/3)=5/(2^(1/3)+3^(1/3))
注：a^2-ab+b^2=(a^3+b^3)/(a+b)
所以只需证2.5<2^(1/3)+3^(1/3)<5
显然2^(1/3)+3^(1/3)<2+3=5
又(5/4)^3=125/64<128/64=2 所以1.25<2^(1/3)
从而2^(1/3)+3^(1/3)>1.25+1.25=2.5
综上原题得证