Question

If t square-4t+ 1=0 then the value of tcube+1/tcube is

Answer 1

if t^2-4t+1=0 then,
solving eq by formula,
comparing eq with ax^2+bx+c=0
we get, a=1,b=-4 ,c=1
b^2-4ac= (-4)^2-4×1×1
=16-4
=12
now t=4+2 root 3/2 ...........formula &-(-)=+
=2(2+root 3)/2
t =2+root 3
t^3=(2+root 3)^3
=8+12 root3 + 18+3 root 3
=26+15 root3
now,t^3+1/t^3=(26+15root3)^2/26+15root3
=26+15root3

t^3+1/t^3=26+15root3.