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Question

Answers

1) $3xyz$

2) ${x^2}yz$

3) $x{y^2}z$

4) $xy{z^2}$

Answer
Verified

We are given that $x + y + z = 0$

We will take $z$ to other side and rewrite the equation as,

$x + y = - z$ (1)

Since, we want to find the value of ${x^3} + {y^3} + {z^3}$ and all of its terms are cubic, hence, take cube on both sides of the equation, $x + y = - z$

Therefore, we have,

${\left( {x + y} \right)^3} = {\left( { - z} \right)^3}$

Now, simplify the equation, using the formula, ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3{a^2}b + 3a{b^2}$

Then, we will get,

${x^3} + {y^3} + 3{x^2}y + 3x{y^2} = - {z^3}$

Keep ${x^3} + {y^3} + {z^3}$ and rest all other terms on one side.

${x^3} + {y^3} + {z^3} = - 3{x^2}y - 3x{y^2}$

Next, we will solve the right hand side of the equation by taking terms common.

${x^3} + {y^3} + {z^3} = - 3xy\left( {x + y} \right)$

From equation (1), we know that $x + y = - z$

Hence, solve the bracket by substituting the value of $x + y$

Therefore, we get,

$

{x^3} + {y^3} + {z^3} = - 3xy\left( { - z} \right) \\

\Rightarrow {x^3} + {y^3} + {z^3} = 3xyz \\

$

Hence, option A is the correct answer.

$

{x^3} + {y^3} + {z^3} - 3xyz = \left( 0 \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right) \\

\Rightarrow {x^3} + {y^3} + {z^3} - 3xyz = 0 \\

\Rightarrow {x^3} + {y^3} + {z^3} = 3xyz \\

$

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