Question

[√(1+C^3)-√(1-C^3)]÷c^2
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Answer 1

[√(1+C^3)-√(1-C^3)]÷c^2 it is possible when [√(1+C^3)-√(1-C^3)]÷c^2= 0

$\frac{ \sqrt{1 + {c}^{3}} - \sqrt{1 - {c}^{3} } }{ {c}^{2} } = 0 \\ = > \sqrt{1 + {c}^{3}} - \sqrt{1 - {c}^{3} } = 0 \\ = > ( { \sqrt{1 + {c}^{3}} - \sqrt{1 - {c}^{3} }})^{2} = {0}^{2} \\ = > {( \sqrt{1 + {c}^{3} } )}^{2} - 2 \times \sqrt{1 + {c}^{3}} \times \sqrt{1 - {c}^{3}} + ({\sqrt{1 - {c}^{3}}})^{2} = 0 \\ = > 1 + {c}^{3} - 2 \sqrt{(1 + {c}^{3})(1 - {c}^{3} ) } + 1 - {c}^{3} = 0 \\ = > 2 - 2 \sqrt{ {1}^{2} - {c}^{3 \times 2} } = 0 \\ = > 2(1 - \sqrt{1 - {c}^{6} } ) = 0 \\ = > 1 - \sqrt{1 - {c}^{6} } = 0 \\ = > - \sqrt{1 - {c}^{6} } = - 1 \\ = > \sqrt{1 - {c}^{6} } = 1 \\ = > {(\sqrt{1 - {c}^{6} }})^{2} = {1}^{2} \\ = > 1 - {c}^{6} = 1 \\ = > - {c}^{6} = 1 - 1\\ = > {c}^{6} = 0 \\ = > c = \sqrt[6]{0} \\ = > c = 0$

Answer 2

Answer:

0 is your answer.

some problem in internet connection because so I can't able to give full answer.

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