α = x(a×b) + y(b×c) + z(c×a) and [a b c]=1/8 then x + y + z =

a·α=0+y[a b c]+0 ⇒ y=8(a·α)Similarly b·α=z18 ⇒ z=8(b·α)x=8 (c·α)x+y+z=8 α·(a+b+c)

α = x(a×b) + y(b×c) + z(c×a) and [a b c]=1/8 then x + y + z =

a·α=0+y[a b c]+0 ⇒ y=8(a·α)Similarly b·α=z18 ⇒ z=8(b·α)x=8 (c·α)x+y+z=8 α·(a+b+c)

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$α = x (a \times b) + y (b \times c) + z (c \times a)$ and $[a b c] = 1 / 8$ then $x + y + z =$

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$8 α . (a + b + c)$

$α . (a + b + c)$

$8 . (a + b + c)$

$a + b + c$

answer is A.

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$a \cdot α = 0 + y [a b c] + 0 \Rightarrow y = 8 (a \cdot α)$

Similarly $b \cdot α = z (\frac{1}{8}) \Rightarrow z = 8 (b \cdot α)$

$\begin{array}{rcl} x & = & 8 (c \cdot α) \\ x + y + z & = & 8 α \cdot (a + b + c) \end{array}$

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α = x(a×b) + y(b×c) + z(c×a) and [a b c]=1/8 then x + y + z =

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$α = x (a \times b) + y (b \times c) + z (c \times a)$ and $[a b c] = 1 / 8$ then $x + y + z =$