If p = 2 – a. Prove that a³ + 6ap + p³ – 8 = 0

Given: p = 2 – a

To Prove: a³ + 6ap + p³ – 8 = 0

Proof:

Substituting the value of p to the LHS expression, we have

a³ + 6a(2 – a) + (2 – a)³ – 8

a³ + 12a – 6a² + 8 – a³ – 6a(2 – a) – 8

Using formula:

(x – y)³ = x³ – y³ – 3xy(x – y)

a³ + 12a – 6a² + 8 – a³ – 12a + 6a² – 8 = 0

Hence proved LHS = RHS