Question
If a + b + c = 15 and a2 + b2 + c2 = 83 then the value of a3 + b3 + c3 - 3abc = ?
Answer: Option B
$$\eqalign{
& a + b + c = 15{\text{ }} \cr
& {a^2} + {b^2} + {c^2} = 83{\text{ }}\left( {{\text{Given}}} \right) \cr
& \therefore a + b + c = 15 \cr
& \left( {{\text{Squaring both sides}}} \right){\text{ }} \cr
& \Rightarrow {\left( {a + b + c} \right)^2} = {\left( {15} \right)^2}{\text{ }} \cr
& \Rightarrow {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = 225 \cr
& \Rightarrow 83 + 2\left( {ab + bc + ca} \right) = 225 \cr
& \Rightarrow 2\left( {ab + bc + ca} \right) = 142 \cr
& \Rightarrow ab + bc + ca = 71 \cr} $$
$$\therefore {a^3} + {b^3} + {c^3} - 3abc = $$ $$\left( {a + b + c} \right)$$ $$\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$$
$$\eqalign{
& \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 15\left( {83 - 71} \right) \cr
& \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 15 \times 12 \cr
& \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 180 \cr} $$
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$$\eqalign{
& a + b + c = 15{\text{ }} \cr
& {a^2} + {b^2} + {c^2} = 83{\text{ }}\left( {{\text{Given}}} \right) \cr
& \therefore a + b + c = 15 \cr
& \left( {{\text{Squaring both sides}}} \right){\text{ }} \cr
& \Rightarrow {\left( {a + b + c} \right)^2} = {\left( {15} \right)^2}{\text{ }} \cr
& \Rightarrow {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca = 225 \cr
& \Rightarrow 83 + 2\left( {ab + bc + ca} \right) = 225 \cr
& \Rightarrow 2\left( {ab + bc + ca} \right) = 142 \cr
& \Rightarrow ab + bc + ca = 71 \cr} $$
$$\therefore {a^3} + {b^3} + {c^3} - 3abc = $$ $$\left( {a + b + c} \right)$$ $$\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)$$
$$\eqalign{
& \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 15\left( {83 - 71} \right) \cr
& \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 15 \times 12 \cr
& \Rightarrow {a^3} + {b^3} + {c^3} - 3abc = 180 \cr} $$
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