Find the value of $a^{3} + b^{3} + c^{3} - 3 a b c$ if $a + b + c = 8$ and $a b + b c + c a = 19$

0

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56

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23

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48

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Solution

The correct option is C $56$
First we find the value of $a^{2} + b^{2} + c^{2}$
$a^{2} + b^{2} + c^{2} = (a + b + c)^{2} - 2 (a b + b c + c a) = 8^{2} - 2 \times 19 = 64 - 38 = 26$
$∴ a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a) = (8) (26 - 19) = 8 \times 7 = 56$

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Similar questions

If a + b + c = 9 and ab + bc + ca = 23, then $a^{3} + b^{3} + c^{3} - 3 a b c =$

a + b + c = 8

a b + b c + c a = 20

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

a + b + c

a b + b c + c a = 26

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

a + b + c = 9

ab + bc + ca = 26,

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

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