Question
If a + b + c = 0, then the value of a3 + b3 + c3 is?
Answer: Option C
$$\eqalign{
& a + b + c = 0 \cr
& {\text{Let, }}{a^3} + {b^3} + {c^3} = T \cr} $$
$$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \frac{1}{2}\left( {a + b + c} \right)$$ $$\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right]$$
$$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( 0 \right)$$ $$\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right]$$
$$\eqalign{
& \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 0 \cr
& \Rightarrow {a^3} + {b^3} + {c^3} = 3abc \cr} $$
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$$\eqalign{
& a + b + c = 0 \cr
& {\text{Let, }}{a^3} + {b^3} + {c^3} = T \cr} $$
$$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \frac{1}{2}\left( {a + b + c} \right)$$ $$\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right]$$
$$ \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = \left( 0 \right)$$ $$\left[ {{{\left( {a - b} \right)}^2} + {{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2}} \right]$$
$$\eqalign{
& \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 0 \cr
& \Rightarrow {a^3} + {b^3} + {c^3} = 3abc \cr} $$
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