Question : If $x^{2}+y^{2}+2x+1=0$, then the value of $x^{31}+y^{35}$ is:
Option 1: –1
Option 2: 0
Option 3: 1
Option 4: 2
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Correct Answer: –1
Solution : Given: $x^{2}+y^{2}+2x+1=0$ ⇒ $x^{2}+2x+1+y^{2}=0$ ⇒ $(x+1)^2+y^2=0$ Since the addition of the squares of the two terms is zero, So, both the terms are individually zero. Therefore, $(x+1)^2=0$ and $y^2=0$. ⇒ $x=-1$ and $y=0$ Now, $x^{31}+y^{35}$ = $(-1)^{31}+(0)^{35}$ = –1 Hence, the correct answer is –1.
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Question : If $(x-5)^2+(y-2)^2+(z-9)^2=0$, then value of $(x+y-z)$ is:
Option 1: 16
Option 2: –1
Option 3: –2
Option 4: 12
Question : If $\frac{(x+y)}{z}=2$, then what is the value of $[\frac{y}{(y-z)}+\frac{x}{(x-z)}]?$
Option 1: $0$
Option 2: $1$
Option 3: $2$
Option 4: $–1$
Question : If $4x–7y=3$ and $2x–y=9$, then the value of $(x–y)$ is:
Option 1: 3
Option 2: 4
Option 3: 5
Option 4: 6
Question : If $x+y+z=0$ and $x^2+y^2+z^2=40$, then what is the value of $x y+y z+z x?$
Option 1: –20
Option 2: 5
Option 3: –5
Option 4: –10
Question : If $x + y = 1$, then what is the value of $x^3+ 3xy + y^3$?
Option 2: 1
Option 3: 0
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