Question

If p = 2 - a ,, prove that a² + 6ap + p³ - 8 = 0

Plz Help....​

Answer 1

Answer

given p=2-a

adding a on both sides we get

a+p=2

given equation is

a³+6ap+p³-8=0

a³+p³+6ap-8=0

a³+b³=(a+b)(a²+b²-ab)

(a+p)(a²+p²-ap)+6ap-8=0

2(a²+p²-ap)+6ap-8=0                    since a+p=2

2a²+2p²-2ap+6ap-8=0

2a²+2p²+4ap-8=0

2(a²+p²+2ap)-8=0

2(a+p)²-8=0

2(2²)-8=0

2(4)-8=0

8-8=0

0=0

hence it is proved a³+6ap+p³-8=0

Step-by-step explanation:

Answer 2

Answer:

p = 2-a

p + a = 2

(p + a)²= 2²

a² + 2ap + p² = 4

a² + p² + 2ap + 4 = 0

a²+p²+6ap =0

hope i help u