Question : If $x= 19$ and $y= 18$, then the value of $\frac{x^{2}+y^{2}+xy}{x^{3}-y^{3}}$ is:
Option 1: 1
Option 2: 37
Option 3: 324
Option 4: 361
Correct Answer: 1
Solution : Given: $x = 19$ and $y= 18$ Since, $x^{3}-y^{3} = (x-y)(x^{2}+y^{2}+xy)$ $\therefore\frac{x^{2}+y^{2}+xy}{x^{3}-y^{3}}=\frac{x^{2}+y^{2}+xy}{ (x-y)(x^{2}+y^{2}+xy)}=\frac{1}{x-y}=\frac{1}{19-18}= 1$ Hence, the correct answer is 1.
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Question : If $(2 x+3 y+4)(2 x+3 y-5)$ is equivalent to $\left(a x^2+b y^2+2 h x y+2 g x+2 f y+c\right)$, then what is the value of $\frac{g + f - c}{abh}$?
Question : If $\frac{x}{y}+\frac{y}{x}=1$ and $x+y=2$, then the value of $x^3+y^3$ is:
Question : If $x+y+z=19, x y z=216$ and $x y+y z+z x=114$, then the value of $x^3+y^3+z^3+x y z$ is:
Question : If $x^4+x^2 y^2+y^4=133$ and $x^2-x y+y^2=7$, then what is the value of $xy$?
Question : If $\frac{x}{4 y}=\frac{3}{4}$ then, the value of $\frac{2 x+3 y}{x–2 y}$ is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile