Question

If a + 2 b + c = 0 then show that a cube plus b cube plus c cube =6abc​

Answer 1

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Answer 2

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$\huge\bold\pink{{solution:-}}$

$we \: know \: that \\ when \: \\ a \: + b \: + c = 0 \\ {a}^{3} + {b}^{3} + {c}^{3} - 3abc = 0 \\ then \\ {a}^{3} + {b}^{3} + {c}^{3} = 3 \: abc \\ here \: given \\ a \: + \: 2b + \: c = 0 \\ \\ here \: , a \: = a \: \\ b \: = 2b \\ and \: \: c = c \\ so \: that \\ {a}^{3} + {b}^{3} + {c}^{3} = 3 \times a \times 2b \: \times \: c \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 6 \: abc$

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