1.

Prove that (a + b)3 + (b + c)3 + (c + a)3 – 3(a + b) (b + c)(c + a) = 2(a3 + b3 + c3 – 3 abc)

Answer»

Let a + b = x, b + c = y and c + a = z 

∴ (a + b)3 + (b + c)3 + (c + a)3 – 3(a + b) (b + c) (c + a) 

= x3 + y3 + z3 – 3xyz 

But x3 + y3 + z3 – 3xyz 

= (x + y + z)(x2 + y2 + z2 – xy – yz – zx) 

=> {(a + b) + (b + c) + (c + a)} {(a + b)2 + (b + c)2 + (c + a)2 – (a + b) (b + c) – (b + c) x (c + a) – (c + a) (a + b)} 

= 2(a + b – c) {a2 + b2 + 2ab + b2 + c2 + 2 bc + c2 + a2 + 2 ca – (ab + ac + b2 + bc) – (bc + ba+ c2 + ac) – (ca + cb + a2 + b2)} 

= 2(a + b + c) (a2 + b2 + c2 – ab – bc – ca) 

= 2(a3 + b3 + c3 – 3 abc)


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