If the centroid of $△$ ABC, in which A(a, b), B(b, c), C(c, a) is at the origin, and abc = 2 then calculate the value of $(a^{3} + b^{3} + c^{3}) .$

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Solution

The correct option is D 6

centroid $= (\frac{x 1 + x 2 + x 3}{3}, \frac{y 1 + y 2 + y 3}{3})$

$(0, 0) = (\frac{a + b + c}{3}, \frac{b + c + a}{3})$
$\frac{a + b + c}{3} = 0$

a + b + c = 0

If a + b + c = 0

then, as we know

$a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - a c)$

$∴ a^{3} + b^{3} + c^{3} - 3 a b c = 0$ … [Since a + b + c = 0]

$∴ a^{3} + b^{3} + c^{3} = 3 a b c = 3 (2) = 6$

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Similar questions

∣ ∣ ∣ ∣ \begin{matrix} a & b & c a - b & b - c & c - a b + c & c + a & a + b \end{matrix} ∣ ∣ ∣ ∣ = a^{3} + b^{3} + c^{3} - 3 a b c

|\begin{matrix} a - b & b + c & a \\ b - c & c + a & b \\ c - a & a + b & c \end{matrix}|

a^{3} + b^{3} + c^{3}

a^{3} + b^{3} + c^{3} - 3 a b c

(\frac{a^{3} + b^{3} + c^{3}}{a b c})

|\begin{matrix} a & b & c \\ a - b & b - c & c - a \\ b + c & c + a & a + b \end{matrix}| = a^{3} + b^{3} + c^{3} - 3 a b c

G

△ A B C

G^{'}

△ A^{'} B^{'} C^{'}

A A^{'} + B B^{'} + C C^{'}

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