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Question

If $$\displaystyle x=a\left ( b-c \right );y=b\left ( c-a \right );z=c\left ( a-b \right )$$, then $$\displaystyle \left ( \frac{x}{a} \right )^{3}+\left ( \frac{y}{b} \right )^{3}+\left ( \frac{z}{c} \right )^{3}$$ is equal to:


A
xyzabc
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B
13xyzabc
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C
3xyzabc
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D
3(x+y+z)(abc)
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Solution

The correct option is C $$\displaystyle 3\frac{xyz}{abc}$$
Let $$\dfrac { x }{ a } =b-c$$$$\dfrac { y }{ b } =c-a$$ and $$\dfrac { z }{ c } =a-b$$.

We know that in general, $$a^3+b^3+c^3=3abc$$ when $$a+b+c=0$$.

Here, $$\dfrac { x }{ a } +\dfrac { y }{ b } +\dfrac { z }{ c } =b-c+c-a+a-b=0$$, therefore,

$$\left( \dfrac { x }{ a }  \right) ^{ 3 }+\left( \dfrac { y }{ b }  \right) ^{ 3 }+\left( \dfrac { z }{ c }  \right) ^{ 3 }=3\left( \dfrac { x }{ a }  \right) \left( \dfrac { y }{ b }  \right) \left( \cfrac { z }{ c }  \right) =3\cfrac { xyz }{ abc }$$.

Hence, $$\left( \cfrac { x }{ a }  \right) ^{ 3 }+\left( \cfrac { y }{ b }  \right) ^{ 3 }+\left( \cfrac { z }{ c }  \right) ^{ 3 }=3\cfrac { xyz }{ abc }$$.

Therefore, option $$C$$ is correct.

Mathematics

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