Question

find odd one out 1, 8, 65, 64

Answer 1

Answer:

$\sqrt[3]{1} = 1$

$\sqrt[3]{8} = 2$

$\sqrt[3]{64} = 4$

$\sqrt[3]{65} = 4.207........$

here,

all are perfect cube roots without 65

so, 65 is the odd number

Answer 2

Answer:

The odd one out is 65 and the correct term should be 27.

Perfect Cubes:

• A perfect cube is a product of three equal integers.
• In other terms, a cube is the product obtained when we multiply a number by itself three times.
• For example, if we multiply 5 three times, we get 125. So, 125 is the cube of 5.
• In the same manner, if we take the cube root of a perfect cube, we get a natural number.
• For example, 216 is a perfect cube because ∛216 = 6.

Step-by-step explanation:

The given series is 1, 8, 65, 64.

• First term: 1 = 1³
• Second term: 8 = 2³
• Third term: 65 = 4.0207³
• Fourth term: 64 = 4³

The third term, i.e. 65 is not a perfect cube as ∛65 = 4.0207

The correct term in the third position should be the cube of three, i.e.,

• Correct third term: 27 = 3³

Therefore, the odd one out from the given series is 65 and the correct term is 27.

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