The cube root of any multiple of $8$ is always divisible by:

2

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4

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8

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16

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Solution

The correct option is A $2$
Prime factorising $8$ , we get,
$8 = 2 \times 2 \times 2 = 2^{3}$ .

If $x$ is a cube and a multiple of $8,$
then $x = 2^{3} \times y$ .

So, cube root of $x$ is divisible by $2.$

Hence, option $A$ is correct.

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\times

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\frac{(36 \times 4) - 8 \times 4}{4 + 8 \times 2 + 16 + 1}

ω

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(1 - ω^{4} + ω^{8}) (1 - ω^{8} + ω^{16})

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$(1 - ω + ω^{2}) (1 - ω^{2} + ω^{4}) (1 - ω^{4} + ω^{8}) (1 - ω^{8} + ω^{16})$ is, where $ω$ is the cube root of unity

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\frac{(36 \times 4) - 8 \times 4}{4 + 8 \times 2 + 16 \div 1}

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