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Inequality

If a1/3+b1/...

Question

If $a^{1 / 3} + b^{1 / 3} + c^{1 / 3} = 0$ , then the value of ${(a + b + c)}^{3}$ will be:

9 a^{2} b^{2} c^{2}

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3 a b c

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6 a b c

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27 a b c

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Solution

The correct option is D $27 a b c$
Since $a^{1 / 3} + b^{1 / 3} + c^{1 / 3} = 0$
${(a^{1 / 3})}^{3} + {(b^{1 / 3})}^{3} + {(c^{1 / 3})}^{3} = 3 a^{1 / 3} b^{1 / 3} c^{1 / 3}$
$\Rightarrow a + b + c = 3 {(a b c)}^{1 / 3}$
$\Rightarrow {(a + b + c)}^{3} = 27 a b c$

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Similar questions

a^{1 / 3} + b^{1 / 3} + c^{1 / 3} = 0

(a + b + c)^{3} = 27 a b c

a_{1} x^{2} + b_{1} x + c_{1} = 0, a_{2} x^{2} + b_{2} x + c_{2} = 0

a_{3} x^{2} + b_{3} x + c_{3} = 0

\frac{c_{1} a_{2} + c_{2} a_{1}}{c_{1} a_{2} - c_{2} a_{1}} + \frac{a_{1} a_{2} b_{3}}{a_{3} (a_{1} b_{2} - a_{2} b_{1})} = 0

{(\frac{a_{1} b_{2} - a_{2} b_{1}}{a_{1} c_{2} - a_{2} c_{1}})}^{2} = \frac{a_{1} a_{2} c_{3}}{c_{1} c_{2} a_{3}}

a^{\frac{1}{3}} + b^{\frac{1}{3}} + c^{\frac{1}{3}} = 0

(a + b + c) = 27 a b c

x

⎡ ⎢ ⎣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ⎤ ⎥ ⎦

¯ ¯ ¯ a = (a_{1}, a_{2}, a_{3}); ¯ ¯ b = (b_{1}, b_{2}, b_{3})

¯ ¯ c = (c_{1}, c_{2}, c_{3})

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