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Evaluate. ^3 sqrt -64 -4 4 undefined
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cube root (like any nth root, for an odd integer n) is never undefined.
only EVENth root can result in 'i' but ODDth root will not.
now, hint: 4^3=64
so its 4?
well... let's see 4 * 4 * 4 = -64 ?
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\(\large \bf \sqrt[3]{-64}\implies \sqrt[3]{-1\cdot 64}\implies \sqrt[3]{-1}\cdot \sqrt[3]{64}\implies \sqrt[3]{(-1)^3}\cdot \sqrt[3]{64} \)
you are close, (-4)^3=-64
(you is you, and jdoe is correct. Not that it takes me to verify that)
so -4
yes
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One more please, can you tell me what will \(\large\color{black}{ \displaystyle \sqrt[3]{-8} }\) be?
2 ^3 sqrt -1?
(-2)^3=(-8) just as 2^3=8
so, \(\large\color{slate}{\displaystyle\sqrt[3]{-8}=-2}\)
ohh ok thnx
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