Question : If $x+y=1+xy$, then $x^3+y^3-x^3y^3$ is equal to:
Option 1: 0
Option 2: 1
Option 3: –1
Option 4: 2
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Correct Answer: 1
Solution : Given: $x+y=1+xy$ (equation 1) We know that the algebraic identity is $(x^3+y^3)=(x+y)^3-3xy(x+y)$ $x^3+y^3-x^3y^3=(x+y)^3-3xy(x+y)-x^3y^3$ Substitute the value from equation 1 in the above equation and we get, $x^3+y^3-x^3y^3=(1+xy)^3-3xy(1+xy)-x^3y^3$ ⇒ $x^3+y^3-x^3y^3=1+x^3y^3+3xy+3x^2y^2-3xy-3x^2y^2-x^3y^3$ ⇒ $x^3+y^3-x^3y^3=1$ Hence, the correct answer is 1.
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Question : If $xy(x+y)=m$, then the value of $(x^3+y^3+3m)$ is:
Option 1: $\frac{m^3}{xy}$
Option 2: $\frac{m^3}{(x+y)^3}$
Option 3: $\frac{m^3}{x^3y^3}$
Option 4: $mx^3y^3$
Question : If $( x^{3}-y^{3}):( x^{2}+xy+y^{2})=5:1$ and $( x^{2}-y^{2}):(x-y)=7:1$, then the ratio $2x:3y$ equals:
Option 1: $4:1$
Option 2: $2:3$
Option 3: $4:3$
Option 4: $3:2$
Question : If $(x+y)^2=xy+1$ and $x^3-y^3=1$, what is the value of $(x-y)$?
Option 1: 1
Option 2: 0
Question : If $x$,$y$ and $z$ are real numbers such that $(x-3)^2+(y-4)^2+(z-5)^2=0$, then $(x+y+z)$ is equal to:
Option 1: –12
Option 3: 8
Option 4: 12
Question : If $xy = -6$ and $x^3+ y^3= 19$ ($x$ and $y$ are integers), then what is the value of $\frac{1}{x^{–1}}+\frac{1}{y^{–1}}$?
Option 1: –1
Option 2: –2
Option 3: 1
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