Question

If x =root7-root5,y=root5-root3 and z=root3-root7,then find the value of xcube+ycube+zcube-2xyz

Answer 1

x=√7-√5, y=√5-√3, z=√3-√7
x³=(√7-√5)³=√7³-3.√7².√5+3.√7.√5²-√5³=7√7-21√5+15√7-5√5=22√7-26√5
y³=(√5-√3)³=√5³-3.√5².√3+3.√5.√3²-√3³=5√5-15√3+9√5-3√3=14√5-18√3
z³=(√3-√7)³=√3³-2.√3².√7+2.√3.√7²-√7³=3√3-6√7+14√3-7√7=17√3-13√7
xyz=(√7-√5)(√5-√3)(√3-√7)
      =(√35-5-√21+√15)(√3-√7)
      =√105-5√3-3√7+3√5-7√5+5√7+7√3-√105
      =2√3+2√7-4√5
∴, x³+y³+z³-2xyz
=22√7-26√5+14√5-18√3+17√3-13√7-2(2√3+2√7-4√5)
=9√7-12√5-√3-4√3-4√7+8√5
=5√7-4√5-5√3