if p = 2- a find a³+ 6ap+p³-8
Answers
Answered by
2
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.....
Answered by
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
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