Are you good at manipulating?

Algebra Level 2

$x=\dfrac {a-b}{a+b} \quad,\quad y=\dfrac {b-c}{b+c}\quad,\quad z=\dfrac {c-a}{c+a}.$

If $x,y$ and $z$ satisfy the equations above, then express $\dfrac {(1+x)(1+y)(1+z)}{(1-x)(1-y)(1-z)}$

in terms of $a,b$ and $c$ .

-1

0

1

None of the above.

abc

(abc)^{2}

1 solution

Surya Prakash
Oct 20, 2015

Since, $x= \dfrac{a-b}{a+b}$ . Applying Componendo and Dividendo to that we get

$\dfrac{1+x}{1-x} = \dfrac{\left(a+b \right)+ \left(a-b \right)}{\left(a+b \right) - \left(a-b \right)} = \dfrac{a}{b}$

Similarly,

$\dfrac{1+y}{1-y} = \dfrac{b}{c}$ $\dfrac{1+z}{1-z} = \dfrac{c}{a}$

So, multiplying all we get,

$\dfrac{\left(1+x \right) \left(1+y \right) \left(1+z \right)}{\left(1-x \right) \left(1-y \right) \left(1-z \right)} = \dfrac{a}{b} \times \dfrac{b}{c} \times \dfrac{c}{a} = \boxed{1}$

Componendo and Dividendo

Moderator note:

Good use of one of my favorite identities.

Is there the possibility that one of $x, y, z$ is equal to 1? What happens then?

You are fast!!I was just writing that same solution :P! Here's an upvote for your speed!

Rohit Udaiwal - 5 years, 7 months ago

He is sunlight , he is the fastest :P

Nihar Mahajan - 5 years, 7 months ago

lol!! Thank you. Nice Problem

Surya Prakash - 5 years, 7 months ago

Good solution.

Upvoted.

Sai Ram - 5 years, 7 months ago

Are you good at manipulating?

1 solution

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