Question

if p = 2- a find a³+ 6ap+p³-8​

Answer 1

Solution:

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

Hope this helps.....

Answer 2

Answer:

0

Step-by-step explanation:

  ( x + y )^3 = x^3 + y^3 + 3xy( x + y )

Here,

⇒ p = 2 - a

⇒ p + a = 2

     Cube on both sides:

⇒ ( p + a )^3 = 2^3

⇒ p^3 + a^3 + 3ap( a + p ) = 8

               a + b = 2

⇒ p^3 + a^3 + 3ap( 2 ) = 8

⇒ p^3 + a^3 + 6ap = 8

⇒ a^3 + 6ap + p^3 - 8 = 0

   Hence the required value of x^3 + 6ap + p^3 - 8 is 0