1.

If p = 2 – a, then prove that a3 + 6ap + p3 – 8 = 0.

Answer»

p = 2 – a => a + p – 2 = 0 

∴ a3 + 6ap + p3 – 8 = (a)3 + (p)3 + (-2)3 – 3a x p x (-2) 

a3 + 6ap + p3 – 8 = {a + p + (-2)} {a2 + p2 + (-2)2 – ap – p(-2) – a(-2)} 

= (a + p – 2) (a2 + p2 + 4 – ap + 2p + 2a) 

= 0 x (a2 + p2 + 4 – ap + 2p + 2a) = 0 

So, a3 + 6ap + p3 – 8 = 0


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