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Question : 33 of 160
Marks:
+1,
-0
Solution:
&(a+2b−c){(a−b)×(a−b−c)}
‌‌=(a+2b−c)[a×a−a×b−a×c−b×a+b×b+b×c]
‌‌=(a+2b−c)[−a×b−a×c−b×a+b×c]
‌‌=(a+2b−c)[b×a+c×a+a×b+b×c]
‌‌=a⋅(b×a)+a⋅(c×a)+a⋅(a×b)
+a⋅(b×c)
+2b⋅(b×a)+2b⋅(c×a)+2b(a×b)
+2b⋅(b×c)
‌‌−c⋅(b×a)−c(c×a)−c(a×b)−c(b×c)
‌‌=[abc]+2[bca]−[cba]−[cab]
‌‌=[abc]+2[abc]+[abc]−[abc]=3[abc]
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