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    Cube Root of βˆ’5

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    What is the cube root of βˆ’5?

    Answer

    βˆ›(βˆ’5) =
    βˆ’βˆ›5 β‰ˆ βˆ’1.71

    Real cube root: the only number whose cube is βˆ’5.

    β‰ˆ β€” rounded to 4 decimal places. βˆ’5 is not the cube of a rational number, so βˆ›(βˆ’5) is irrational: its decimal expansion never ends and never repeats.

    A negative number has a negative cube root: (βˆ’βˆ›5)Β³ = βˆ’5. There is no real square root of βˆ’5.

    Sign
    βˆ›(βˆ’5) = βˆ’βˆ›5, because (βˆ’a)Β³ = βˆ’aΒ³
    Prime factors
    5 is a prime number: no triple of equal factors.
    Between cubes
    1Β³ = 1 < 5 < 8 = 2Β³, so 1 < βˆ›5 < 2 and βˆ›5 β‰ˆ 1.71

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    https://calculat.io/en/number/cube-root/-5Copy
    <a href="https://calculat.io/en/number/cube-root/-5">Cube Root of βˆ’5 (βˆ›(βˆ’5)): Exact Value and Decimal</a>Copy
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    Cube Root of βˆ’5 (βˆ›(βˆ’5)): Exact Value and Decimal. βˆ›(βˆ’5) = βˆ’βˆ›5 β‰ˆ βˆ’1.71. 5 has no cube factor but 1, so βˆ›5 is already simplest; it lies between 1 and 2. The minus comes out of the root.
    https://calculat.io/en/number/cube-root/-5/generated.jpgCopy

    FAQ

    What is the cube root of βˆ’5?

    βˆ›(βˆ’5) = βˆ’βˆ›5 β‰ˆ βˆ’1.71

    Why does the cube root of βˆ’5 exist but the square root does not?

    A cube keeps the sign: (βˆ’βˆ›5)Β³ = βˆ’5, so βˆ›(βˆ’5) = βˆ’βˆ›5. The square of any real number is never negative, so no real number squared gives βˆ’5.

    Is βˆ›(βˆ’5) a rational number?

    No: βˆ’5 is not the cube of a rational number, and the cube root of a fraction p/q in lowest terms is rational only when p and q are both cubes. So βˆ›(βˆ’5) is irrational; its decimal never ends and never repeats: βˆ›(βˆ’5) β‰ˆ βˆ’1.71.