(a+b) cdot(b+c) times(a+b+c) is equal to | vedclass

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(B) We are given the expression $(a+b) \cdot ((b+c) 	imes (a+b+c))$.First,expand the cross product term: $(b+c) 	imes (a+b+c) = (b 	imes a) + (b

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$[a b c]$

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(B) We are given the expression $(a+b) \cdot ((b+c) 	imes (a+b+c))$.First,expand the cross product term: $(b+c) 	imes (a+b+c) = (b 	imes a) + (b 	imes b) + (b 	imes c) + (c 	imes a) + (c 	imes b) + (c 	imes c)$.Since the cross product of a vector with itself is zero ($b 	imes b = 0$ and $c 	imes c = 0$) and $c 	imes b = -(b 	imes c)$,we have:$(b+c) 	imes (a+b+c) = (b 	imes a) + (b 	imes c) + (c 	imes a) - (b 	imes c) = (b 	imes a) + (c 	imes a)$.Now,take the dot product with $(a+b)$:$(a+b) \cdot ((b 	imes a) + (c 	imes a)) = a \cdot (b 	imes a) + a \cdot (c 	imes a) + b \cdot (b 	imes a) + b \cdot (c 	imes a)$.Using the properties of the scalar triple product $[x y z] = x \cdot (y 	imes z)$:$a \cdot (b 	imes a) = [a b a] = 0$ (since two vectors are same).$a \cdot (c 	imes a) = [a c a] = 0$.$b \cdot (b 	imes a) = [b b a] = 0$.$b \cdot (c 	imes a) = [b c a]$.Since $[b c a] = [a b c]$,the final result is $[a b c]$.

$(a+b) \cdot(b+c) \times(a+b+c)$ is equal to

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