x² - x - 1 = 0

$$ \therefore $$ $$\alpha $$² - $$\alpha $$ - 1 = 0

$$ \Rightarrow $$ $$\alpha $$² = $$\alpha $$ + 1

$$ \therefore $$ $$\alpha $$³ = $$\alpha $$² + $$\alpha $$

= $$\alpha $$ + 1 + $$\alpha $$

= 2$$\alpha $$ + 1

Now $$\alpha $$⁴ = 2$$\alpha $$² + $$\alpha $$

= 2($$\alpha $$ + 1) + $$\alpha $$

= 3$$\alpha $$ + 2

Now $$\alpha $$⁵ = 3$$\alpha $$² + 2$$\alpha $$

= 3($$\alpha $$ + 1) + 2$$\alpha $$

= 5$$\alpha $$ + 3

Given p_k = $${\left( \alpha \right)^k} + {\left( \beta \right)^k}$$

$$ \therefore $$ p₅ = $${\left( \alpha \right)^5} + {\left( \beta \right)^5}$$

= 5$$\alpha $$ + 3 + 5$$\beta $$ + 3

= 5($$\alpha $$ + $$\beta $$) + 6

= 5(1) + 6 [As $$\alpha $$ + $$\beta $$ = 1]

= 11

Now p₂ · p₃

= ($$\alpha $$² + $$\beta $$²).($$\alpha $$³ + $$\beta $$³)

= ( $$\alpha $$ + 1 + $$\beta $$ + 1)(2$$\alpha $$ + 1 + 2$$\beta $$ + 1)

= ( $$\alpha $$ + $$\beta $$ + 2)(2($$\alpha $$ + $$\beta $$) + 2)

= (1 + 2)(2 + 2) = 12

$$ \therefore $$ p₅ $$ \ne $$ p₂ ·p₃

So option (D) is wrong.