Question : If $x+y+z=1, \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1,$ and $xyz=-1$, then $x^3+y^3+z^3 $ is equal to:
Option 1: –1
Option 2: 1
Option 3: –2
Option 4: 2
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 1
Solution : Given: $x+y+z=1,\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$, $xyz=-1$ Consider, $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$ ⇒ $xy+yz+zx=xyz=-1$ Now, $x+y+z=1$ ⇒ $(x+y+z)^2=1^2$ ⇒ $x^2+y^2+z^2+2(xy+yz+zx)=1$ ⇒ $x^2+y^2+z^2+2(-1)=1$ ⇒ $x^2+y^2+z^2=3$ We know, $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-(xy+yz+zx)$ So, putting all the values, we get, ⇒ $x^3+y^3+z^3-3×(-1)=1×[3-(-1)]$ ⇒ $x^3+y^3+z^3=4-3$ $\therefore x^3+y^3+z^3=1$ Hence, the correct answer is 1.
Candidates can download this ebook to know all about SSC CGL.
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Question : If $\frac{(x+y)}{z}=2$, then what is the value of $[\frac{y}{(y-z)}+\frac{x}{(x-z)}]?$
Question : The value of $\frac{(x-y)^3+(y-z)^3+(z-x)^3}{6(x-y)(y-z)(z-x)}$, where $x \neq y \neq z$, is equal to:
Question : If $(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) = 0$ and $x+y+z = 9$, what is the value of $x^{3} + y^{3}+z^{3}-xyz$?
Question : If $x+y+z=0$, then what is the value of $\frac{x^2}{(y z)}+\frac{y^2}{(x z)}+\frac{z^2}{(x y)}$?
Question : $\text { If } x^2+y^2+z^2=x y+y z+z x \text { and } x=1 \text {, then find the value of } \frac{10 x^4+5 y^4+7 z^4}{13 x^2 y^2+6 y^2 z^2+3 z^2 x^2}$.
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile