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

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(B) We need to evaluate the scalar triple product: $(a+b) \cdot ((b+c) 	imes (a+b+c))$.First, expand the cross product term: $(b+c) 	imes (a+b+c) =

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$[a b c]$

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(B) We need to evaluate the scalar triple product: $(a+b) \cdot ((b+c) 	imes (a+b+c))$.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 any vector with itself is zero ($b 	imes b = 0$ and $c 	imes c = 0$), and using the property $c 	imes b = -(b 	imes c)$, we get:$(b+c) 	imes (a+b+c) = (b 	imes a) + (b 	imes c) + (c 	imes a) - (b 	imes c) = (b 	imes a) + (c 	imes a)$.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 scalar triple products, $a \cdot (b 	imes a) = 0$, $a \cdot (c 	imes a) = 0$, and $b \cdot (b 	imes a) = 0$.This leaves us with $b \cdot (c 	imes a) = [b c a]$.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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