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
Correct Answer: 1225
Solution : $x + y + z = 19$ --------------(1) $xyz = 216$ --------------------(2) $xy + yz + zx= 114$ -----------(3) Squaring both sides in equation (1) $(x + y + z)^2 = 19^2$ $⇒x^2 + y^2 + z^2 + 2(xy + yz + zx) = 361$ $⇒x^2 + y^2 + z^2 = 361 - 2 × 114$ $⇒x^2 + y^2 + z^2 = 133$ Now, $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$ $⇒x^3 + y^3 + z^3= 19(133 - 114) + 648$ $⇒x^3 + y^3 + z^3 = 19 × 19 + 648$ $⇒x^3 + y^3 + z^3= 1009$ So, $x^3 + y^3 + z^3 + xyz= 1009 + 216= 1225$ Hence, the correct answer is 1225.
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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 $x+y+z=10$, $x y+y z+z x=25$ 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
Question : If $x+y+z=13,x^2+y^2+z^2=133$ and $x^3+y^3+z^3=847$, then the value of $\sqrt[3]{x y z}$ is:
Option 1: $8$
Option 2: $7$
Option 3: $-9$
Option 4: $-6$
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
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