QUESTION

# If $a + b + c = 0$then ${a^3} + {b^3} + {c^3}$ is equal toA. $3abc$B. $\dfrac{3}{{abc}}$C. $3{a^3}{b^3}{c^3}$D. Zero

Hint: In this question use the algebraic formula ${a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - bc - ca - ab} \right)$
$a + b + c = 0$
As we know that ${a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - bc - ca - ab} \right)$
${a^3} + {b^3} + {c^3} - 3abc = 0 \times \left( {{a^2} + {b^2} + {c^2} - bc - ca - ab} \right) \\ \therefore {a^3} + {b^3} + {c^3} - 3abc = 0 \\ \therefore {a^3} + {b^3} + {c^3} = 3abc \\$
Note: In this question we put the value of given equation $a + b + c = 0$ in the cubic algebraic formula and formed the equation after that we simplified it and got the value of ${a^3} + {b^3} + {c^3}$as $3abc$.