If (a + 1/a )2 = 3, then find a18 + a12 + a6 + 1.

The correct answer is Option 1: 0

(a + 1/a)² = 3

Step 1

a + 1/a = ±√3

Step 2

Use identity

a² + 1/a² = (a + 1/a)² − 2

= 3 − 2

= 1

Step 3

Use identity

a³ + 1/a³ = (a + 1/a)³ − 3(a + 1/a)

= (±√3)³ − 3(±√3)

= ±3√3 − ±3√3

= 0

So

a³ + 1/a³ = 0

Step 4

Multiply by a³

a⁶ + 1 = 0

Therefore

a⁶ = −1

Now;

⇒ a¹⁸ + a¹² + a⁶ + 1

= (a⁶)³ + (a⁶)² + a⁶ + 1

= - 1 + 1 - 1 + 1 = 0