Question : If $x + y + z =1$, $xy + yz + zx = - 26$, and $x^3+y^3+z^3 = 151$. then what will be the value of $xyz$?
Option 1: –18
Option 2: 32
Option 3: 24
Option 4: –30
Correct Answer: 24
Solution : Given: $x + y + z =1$, $xy + yz + zx = - 26$, and $x^3+y^3+z^3 = 151$ We know, $x^{2} + y^{2} + z^{2} = (x + y + z)^{2} − 2(xy + yz + zx)$ $\therefore x^{2} + y^{2} + z^{2} = (1)^{2} - 2(-26)= 53$ Now, $x^{3} + y^{3} + z^{3} − 3 xyz = (x + y + z )(x^{2} + y^{2} + z^{2} − xy − yz − zx)$ ⇒ $151 − 3xyz = 1(x^{2} + y^{2} + z^{2} − xy − yz − zx)$ ⇒ $151 − 3xyz = 1(53 − (− 26))$ ⇒ $151− 3xyz = 1(79) = 79$ ⇒ $3xyz = 151 - 79 = 72$ $\therefore xyz = \frac{72}{3}= 24$ Hence, the correct answer is 24.
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Question : If $x+y+z=19, x y z=216$ and $x y+y z+z x=114$, then the value of $\sqrt{x^3+y^3+z^3+x y z}$ is:
Option 1: 32
Option 2: 30
Option 3: 28
Option 4: 35
Question : If $x+y+z=17, x y z=171$ and $x y+y z+z x=111$, then the value of $\sqrt[3]{\left(x^3+y^3+z^3+x y z\right)}$ is:
Option 1: –64
Option 2: 4
Option 3: 0
Option 4: –4
Question : If $x^2 = y+z$, $y^2=z+x$, $z^2=x+y$, then the value of $\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}$ is:
Option 1: –1
Option 2: 1
Option 3: 2
Option 4: 4
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:
Option 1: 1225
Option 2: 1441
Option 3: 361
Option 4: 577
Question : If $x+y+z=10$, $x y+y z+z x=25$ and $x y z=100$, then what is the value of $(x^3+y^3+z^3)$?
Option 1: 450
Option 2: 540
Option 3: 550
Option 4: 570
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