Explanation:

x^2–4x+1 = 0
or x^2+1 = 4x……………..(1)
x^4+2x^2+1= 16x^2
x^4 = 14x^2–1……………..(2)
Now
x^9+x^7–194x^5–194x^3 =?
=x^3[x^6+x^4–194x^2–194]
=x^3[x^4(x^2+1)-194(x^2+1)]
=x^3(x^2+1) (x^4–194)
On putting x^2+1 =4x from eq.(1)
=x^3.(4x)(x^4–194)
=4.x^4(x^4–194)
On putting x^4 =14x^2–1 fromeq.(2)
=4.(14x^2–1) (14x^2–1–194)
=4(14x^2–1) (14x^2–195)
=4.{14(x^2+1)-15}.{14(x^2+1)-209}
On putting (x^2+1)=4x
=4.(56x-15)(56x-209)
=4[(56x)^2 -224 (56x)+3135].
=4.[(56)^2.x^2–4×(56)^2.x+3135]
on putting x^2 =(4x-1)
=4.[(56)^2(4x-1)-4×(56)^2.x+3135]
=4.[4×(56)^2.x-(56)^2–4×(56)^2.x+3135]
=4.[3135-(56)^2]
=4.[3135–3136)
=4.(-1)
= -4