x=3−2√2
1x=13−2√2
1x=3+2√2(3−2√2)(3+2√2)
=3+2√232−(2√2)2
=3+2√29−8
=3+2√2
Now,
x−1x=3−2√2−(3+2√2)
=−4√2
We know,
x2−1x2=(x+1x)2−2(x)(1x)
=(3−2√2+3+2√2)2−2
=62−2
=36−2
=34
Now,
x3−1x3=(x−1x)[x2+1x2+(x)(1x)]
=(−4√2)(34+1)
=(−4√2)35
=−140√2