State True or False.x3+y3+z3−3xyz=12(x+y+z)[(x−y)2+(y−z)2+(z−x)2]
A
True
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B
False
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Solution
The correct option is A True R.H.S=12(x+y+z)[(x−y)2+(y−z)2+(z−x)2] =12(x+y+z)[(x2−2xy+y2)+(y2−2yz+z2)+(z2−2zx+x2) =12(x+y+z)[2x2+2y2+2z2−2xy−2yz−2zx] =12(x+y+z)2(x2+y2+z2−xy−yz−zx] =(x+y+z)(x2+y2+z2−xy−yz−zx) =x3+y3+z3−3xyz =L.H.S