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If sin−1x+sin−1y+sin−1z=π then x4+y4+z4+4x2y2z2=k(x2y2+y2z2+z2x2) where k is equal to |
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Answer» If sin−1x+sin−1y+sin−1z=π then x4+y4+z4+4x2y2z2=k(x2y2+y2z2+z2x2) where k is equal to |
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