Question

if p = 2 - a , Prove that a³ + 6ap + p³ = 8​

Answer 1

Step-by-step explanation:

Given:

• p = 2 - a

To prove:

• a³ + 6ap + p³ = 8

Solution :

We can also write p = 2 - a, as p + a = 2 (By transposing a on LHS).

⇒ p + a = 2

Cube both the sides,

⇒ (p + a)³ = (2)³

We know that, (a + b)³ = a³ + b³ + 3ab(a + b).

⇒ p³ + a³ + 3pa(p + a) = 8

Now, we know the value of (p + a) = 2. Just substitute the value !

⇒ p³ + a³ + 3pa(2) = 8

⇒ p³ + a³ + 6pa = 8

Hence, proved !

Answer 2

Answer:

$p = 2 - a \\ a + p = 2 \\ {(a + p)}^{3} = {2}^{3} \\ {a}^{3} + 3ap(a + p) + {p}^{3} = 8 \\ {a}^{3} + 3ap(2) + {p}^{3} = 8 \\ \boxed{{a}^{3} + 6ap + {p}^{3} = 8}$