Question : If $x=5$ ; $y=6$ and $z=-11$, then the value of $x^{3}+y^{3}+z^{3}$ is:
Option 1: –890
Option 2: –970
Option 3: –870
Option 4: –990
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Correct Answer: –990
Solution : Given: $x=5$, $y=6$ and $z=-11$ Here $x+y+z= 5 + 6 -11 = 0$ So, $x^{3}+y^{3}+z^{3} = 3xyz = 3\times 5\times 6 \times (-11) = -990$ Hence, the correct answer is $-990$.
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Question : If ${x^2+y^2+z^2=2(x+z-1)}$, then the value of $x^3+y^3+z^3$ is equal to:
Option 1: 6
Option 2: 1
Option 3: 2
Option 4: 8
Question : If $x(x+y+z)=20$, $y(x+y+z)=30$, and $z(x+y+z)=50$, then the value of $2(x+y+z)$ is:
Option 1: 20
Option 2: –10
Option 3: 15
Option 4: 18
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 $X$ is 20% less than $Y$, then find the values of$\frac{Y–X}{Y}$ and $\frac{X}{X–Y}$.
Option 1: $\frac{1}{5}$ and $-4$
Option 2: $5$ and $-\frac{1}{4}$
Option 3: $\frac{2}{5}$ and $-\frac{5}{2}$
Option 4: $\frac{3}{5}$ and $-\frac{5}{3}$
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: $\frac{1}{2}(x+y+z)$
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