Prove That :

(a+b)3+ (b+c)3+ (c+a)3- 3(a+b)(b+c)(c+a) = 2(a3+b3+c3-3abc)

We know that x3+y3+z3-3xyz=x+y+z x2+y2+z2-xy-yz-zx.....1Now putting x = a+b, y = b+c, z = c+aa+b3+b+c3+c+a3-3a+b b+cc+a=a+b+b+c+c+aa+b2+b+c2+c+a2-ab-ac-b2-bc-bc-ab-c2-ac-ac-a2-bc-ab=2a+b+c2a2+2b2+2c2+2ab+2bc+2ac-3ab-3bc-3ac-b2-c2-a2=2a+b+ca2+b2+c2-ab-bc-ca=2a3+b3+c3-3abc  Using equation 1

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