Question

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

Answer 1

Hint: In this question use the algebraic formula ${a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - bc - ca - ab} \right)$

Complete step-by-step answer:
Given that,
$a + b + c = 0$
As we know that ${a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - bc - ca - ab} \right)$
Put the given value we get:
$
  {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 \\
$
Hence option A is correct.

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 $3abc$.