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: $mx^3y^3$
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Correct Answer: $\frac{m^3}{x^3y^3}$
Solution : Given: $xy(x+y)=m$ We know that the algebraic identity is $(x+y)^3=x^3+y^3+3xy(x+y)$. $xy(x+y)=m$ ⇒ $(x+y)=\frac{m}{xy}$ Take the cube on both sides of the above equation, we get, $(x+y)^3=(\frac{m}{xy})^3$ ⇒ $x^3+y^3+3xy(x+y)=\frac{m^3}{x^3y^3}$ Substitute the value of $xy(x+y)=m$ in above equation, we get, $x^3+y^3+3m=\frac{m^3}{x^3y^3}$ Hence, the correct answer is $\frac{m^3}{x^3y^3}$.
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Question : If $x=\frac{\sqrt{5}+1}{\sqrt{5}-1}$ and $y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$, then the value of $\frac{x^{2}+xy+y^{2}}{x^{2}-xy+y^{2}}$ is:
Question : If $x^2-xy+y^2=2$ and $x^4+x^2y^2+y^4=6$, then the value of $(x^2+xy+y^2)$ is:
Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{yz}+\frac{y^2}{xz}+\frac{z^2}{xy}$?
Question : If $\frac{x^2}{yz}+\frac{y^2}{zx}+\frac{z^2}{xy}=3$, then what is the value of $(x+y+z)^3$?
Question : If $x^2+y^2=29$ and $xy=10$, where $x>0,y>0$ and $x>y$. Then the value of $\frac{x+y}{x-y}$ is:
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