verify :(9 xyz) {1/3xyz}²=x²y³z³,for x=2,y=1,z=3
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We have, L.H.S.=x3+y3+z3−3xyz
=(x+y+z)(x2+y2+z2−xy−yz−zx) [by poynomial identity]
=21(x+y+z)(2x2+2y2+2z2−2xy−2yz−2zx)
=21[(x+y+z)(x−y)2+(y−z)2+(z−x)2]
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