Question : If $\frac{x}{xa+yb+zc}=\frac{y}{ya+zb+xc}=\frac{z}{za+xb+yc}$ and $x+y+z\neq 0$, then each ratio is:
Option 1: $\frac{1}{a-b-c}$
Option 2: $\frac{1}{a+b-c}$
Option 3: $\frac{1}{a-b+c}$
Option 4: $\frac{1}{a+b+c}$
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Correct Answer: $\frac{1}{a+b+c}$
Solution : Given: $\frac{x}{xa+yb+zc}=\frac{y}{ya+zb+xc}=\frac{z}{za+xb+yc}$ Let $\frac{x}{xa+yb+zc}=\frac{y}{ya+zb+xc}=\frac{z}{za+xb+yc}$ = k Or, $\frac{x+y+z}{xa+yb+zc+ya+zb+xc+za+xb+yc}$ = k Or, $\frac{x+y+z}{y(a+b+c)+x(a+b+c)+z(a+b+c)}$ = k Or, $\frac{x+y+z}{(a+b+c)(x+y+z)}$ = k $\therefore$ k = $\frac{1}{(a+b+c)}$ Hence, the correct answer is $\frac{1}{(a+b+c)}$.
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Question : Simplify the given expression. $\frac{x^3+y^3+z^3-3 x y z}{(x-y)^2+(y-z)^2+(z-x)^2}$
Option 1: $\frac{1}{3}(x+y+z)$
Option 2: $(x+y+z)$
Option 3: $\frac{1}{4}(x+y+z)$
Option 4: $\frac{1}{2}(x+y+z)$
Question : If $\small x=a\left (b-c \right),\; y=b\left (c-a \right) ,\; z=c\left (a-b \right)$, then the value of $\left (\frac{x}{a} \right)^{3}+\left (\frac{y}{b} \right)^{3}+\left (\frac{z}{c} \right)^{3}$ is:
Option 1: $\frac{2xyz}{abc}$
Option 2: $\frac{xyz}{abc}$
Option 3: $0$
Option 4: $\frac{3xyz}{abc}$
Question : If $6^{x}=3^{y}=2^{z}$, then what is the value of $\frac{1}{y}+\frac{1}{z}-\frac{1}{x}$?
Option 1: 1
Option 2: 0
Option 3: 3
Option 4: 6
Question : If $x: y: z=3: 4: 5$, then what will be the ratio of $\left(\frac{x}{y}\right):\left(\frac{y}{z}\right):\left(\frac{z}{x}\right)$?
Option 1: 49 : 37 : 100
Option 2: 45 : 48 : 100
Option 3: 41 : 37 : 100
Option 4: 37 : 47 : 100
Question : If $x$ = $y$ = $z$, then $\frac{\left (x+y+z \right )^{2}}{x^{2}+y^{2}+z^{2}}$ is equal to:
Option 1: 4
Option 2: 2
Option 4: 1
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