Question:
The number of solutions of the equations
\[
x+y+z=12,\quad x^2+y^2+z^2=50,\quad x^3+y^3+z^3=216
\]
is
The number of solutions of the equations
\[
x+y+z=12,\quad x^2+y^2+z^2=50,\quad x^3+y^3+z^3=216
\]
is
Show Hint
When \(x+y+z\), \(x^2+y^2+z^2\), and \(x^3+y^3+z^3\) are given, first find \(xy+yz+zx\) and \(xyz\), then form the cubic equation whose roots are \(x,y,z\).
Updated On: Jun 25, 2026
- \(6\)
- \(24\)
- \(3\)
- \(9\)
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The Correct Option is A
Solution and Explanation
Step 1: Use the given sum of variables.
Given \[ x+y+z=12 \] Let \[ x+y+z=s_1=12 \] Also, \[ x^2+y^2+z^2=50 \] Using the identity \[ x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx), \] we get \[ 50=12^2-2(xy+yz+zx) \] So, \[ 50=144-2(xy+yz+zx) \] Therefore, \[ 2(xy+yz+zx)=94 \] Hence, \[ xy+yz+zx=47 \]
Step 2: Use the sum of cubes identity.
Given \[ x^3+y^3+z^3=216 \] Using the identity \[ x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx), \] we get \[ 216-3xyz=12(50-47) \] So, \[ 216-3xyz=12(3) \] Therefore, \[ 216-3xyz=36 \] Hence, \[ 3xyz=180 \] So, \[ xyz=60 \]
Step 3: Form the cubic equation.
Since \(x,y,z\) are roots of the cubic equation, \[ t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz=0 \] Substituting the values, \[ t^3-12t^2+47t-60=0 \] Now factorizing, \[ t^3-12t^2+47t-60=(t-3)(t-4)(t-5) \] Thus, \[ t=3,4,5 \] So, \[ x,y,z \] are \(3,4,5\) in some order.
Step 4: Count the number of solutions.
Since \(3,4,5\) are distinct values, the number of ordered triples is \[ 3!=6 \]
Step 5: Final conclusion.
Therefore, the number of solutions is \[ \boxed{6} \]
Given \[ x+y+z=12 \] Let \[ x+y+z=s_1=12 \] Also, \[ x^2+y^2+z^2=50 \] Using the identity \[ x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx), \] we get \[ 50=12^2-2(xy+yz+zx) \] So, \[ 50=144-2(xy+yz+zx) \] Therefore, \[ 2(xy+yz+zx)=94 \] Hence, \[ xy+yz+zx=47 \]
Step 2: Use the sum of cubes identity.
Given \[ x^3+y^3+z^3=216 \] Using the identity \[ x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx), \] we get \[ 216-3xyz=12(50-47) \] So, \[ 216-3xyz=12(3) \] Therefore, \[ 216-3xyz=36 \] Hence, \[ 3xyz=180 \] So, \[ xyz=60 \]
Step 3: Form the cubic equation.
Since \(x,y,z\) are roots of the cubic equation, \[ t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz=0 \] Substituting the values, \[ t^3-12t^2+47t-60=0 \] Now factorizing, \[ t^3-12t^2+47t-60=(t-3)(t-4)(t-5) \] Thus, \[ t=3,4,5 \] So, \[ x,y,z \] are \(3,4,5\) in some order.
Step 4: Count the number of solutions.
Since \(3,4,5\) are distinct values, the number of ordered triples is \[ 3!=6 \]
Step 5: Final conclusion.
Therefore, the number of solutions is \[ \boxed{6} \]
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