Question

IF p = 2-a, prove that a³+6ap+p³-8=0

Answer 1

p = 2-a or a+p = 2
a³ + 6ap+p³ -8 = (a³+p³) + 6ap - 8 = (a+p)(a²+p²-ap) + 6ap - 8
= 2(a²+p²-ap) + 6ap - 8 = 2(a²+p²-ap + 3ap) - 8 = 2(a+p)² - 8 = 2x2² - 8 = 0

Answer 2

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