Question:
If $x=\alpha$, $y=\beta$, $z=\gamma$ satisfy the equations
\[
3x+y+2z+2=0,
\]
\[
2x-3y+z-7=0,
\]
\[
x-4y+3z-1=0,
\]
then
\[
\alpha^3-\beta^3=
\]
If $x=\alpha$, $y=\beta$, $z=\gamma$ satisfy the equations
\[
3x+y+2z+2=0,
\]
\[
2x-3y+z-7=0,
\]
\[
x-4y+3z-1=0,
\]
then
\[
\alpha^3-\beta^3=
\]
Show Hint
While solving simultaneous equations, eliminate variables systematically to reduce calculation mistakes.
Updated On: Jun 17, 2026
- $19$
- $-35$
- $0$
- $16$
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
A system of simultaneous linear equations can be solved using elimination or substitution methods.
After obtaining the values of variables, substitute them carefully into the required algebraic expression.
Step 1: Rewrite the equations in simplified form.
\[ 3x+y+2z=-2 \] \[ 2x-3y+z=7 \] \[ x-4y+3z=1 \]
Step 2: Eliminate one variable.
Multiply the third equation by $2$: \[ 2x-8y+6z=2 \] Now subtract the second equation: \[ (2x-8y+6z)-(2x-3y+z)=2-7 \] \[ -5y+5z=-5 \] \[ z-y=-1 \] \[ z=y-1 \]
Step 3: Substitute into the second equation.
Substitute \[ z=y-1 \] into \[ 2x-3y+z=7 \] \[ 2x-3y+(y-1)=7 \] \[ 2x-2y=8 \] \[ x-y=4 \] \[ x=y+4 \]
Step 4: Find the value of $y$.
Substitute \[ x=y+4 \] and \[ z=y-1 \] into the first equation: \[ 3(y+4)+y+2(y-1)=-2 \] \[ 3y+12+y+2y-2=-2 \] \[ 6y+10=-2 \] \[ 6y=-12 \] \[ y=-2 \] Hence, \[ x=2 \] and \[ z=-3 \] Thus, \[ \alpha=2, \qquad \beta=-2 \]
Step 5: Evaluate the required expression.
\[ \alpha^3-\beta^3 = 2^3-(-2)^3 \] \[ =8+8 \] \[ =16 \] Hence, \[ \boxed{16} \]
Step 1: Rewrite the equations in simplified form.
\[ 3x+y+2z=-2 \] \[ 2x-3y+z=7 \] \[ x-4y+3z=1 \]
Step 2: Eliminate one variable.
Multiply the third equation by $2$: \[ 2x-8y+6z=2 \] Now subtract the second equation: \[ (2x-8y+6z)-(2x-3y+z)=2-7 \] \[ -5y+5z=-5 \] \[ z-y=-1 \] \[ z=y-1 \]
Step 3: Substitute into the second equation.
Substitute \[ z=y-1 \] into \[ 2x-3y+z=7 \] \[ 2x-3y+(y-1)=7 \] \[ 2x-2y=8 \] \[ x-y=4 \] \[ x=y+4 \]
Step 4: Find the value of $y$.
Substitute \[ x=y+4 \] and \[ z=y-1 \] into the first equation: \[ 3(y+4)+y+2(y-1)=-2 \] \[ 3y+12+y+2y-2=-2 \] \[ 6y+10=-2 \] \[ 6y=-12 \] \[ y=-2 \] Hence, \[ x=2 \] and \[ z=-3 \] Thus, \[ \alpha=2, \qquad \beta=-2 \]
Step 5: Evaluate the required expression.
\[ \alpha^3-\beta^3 = 2^3-(-2)^3 \] \[ =8+8 \] \[ =16 \] Hence, \[ \boxed{16} \]
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