Question : What is the simplified value of $\frac{(x+y+z)(x y+y z+z x)–x y z}{(x+y)(y+z)(z+x)}$?
Option 1: $y$
Option 2: $z$
Option 3: $1$
Option 4: $x$
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Correct Answer: $1$
Solution :
Given: The expression is $\frac{(x+y+z)(x y+y z+z x)–x y z}{(x+y)(y+z)(z+x)}$.
$\frac{(x+y+z)(x y+y z+z x)–x y z}{(x+y)(y+z)(z+x)}$
Substitute the value of $x=y=z=1$, in the given expression, we get,
= $\frac{(1+1+1)(1\times1+1\times 1+1\times 1)–1\times1\times1}{(1+1)(1+1)(1+1)}$
= $\frac{3\times 3–1}{2\times 2\times 2}=\frac{9–1}{8}$
= $\frac{8}{8}$ = $1$
Hence, the correct answer is 1.
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