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# If $x + y + z = 0$, then ${x^3} + {y^3} + {z^3} = \_\_\_\_\_\_\_$1) $3xyz$2) ${x^2}yz$3) $x{y^2}z$4) $xy{z^2}$  Hint: We will first rearrange the given equation $x + y + z = 0$, by taking $z$ to the other side. Then, take a cube on both sides. Apply the formula of ${\left( {a + b} \right)^3}$ to simplify the equation. Next, solve the equation and find the value of ${x^3} + {y^3} + {z^3}$

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.

Note: Students should remember the basic algebraic identities. The question can also be done using the formula, ${a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$ such as,${x^3} + {y^3} + {z^3} - 3xyz = \left( {x + y + z} \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right)$ and then, substituting the value $x + y + z = 0$ which will give
${x^3} + {y^3} + {z^3} - 3xyz = \left( 0 \right)\left( {{x^2} + {y^2} + {z^2} - xy - yz - zx} \right) \\ \Rightarrow {x^3} + {y^3} + {z^3} - 3xyz = 0 \\ \Rightarrow {x^3} + {y^3} + {z^3} = 3xyz \\$

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