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Making use of the cube root table, find the cube root 1346.

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Question

Making use of the cube root table, find the cube root
1346.

Sum
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Solution

By prime factorisation, we have:

\[1346 = 2 \times 673 \Rightarrow \sqrt[3]{1346} = \sqrt[3]{2} \times \sqrt[3]{673}\]

Also

\[670 < 673 < 680 \Rightarrow \sqrt[3]{670} < \sqrt[3]{673} < \sqrt[3]{680}\]

From the cube root table, we have:

\[\sqrt[3]{670} = 8 . 750 \text{ and }  \sqrt[3]{680} = 8 . 794\]

For the difference (680 - 670), i.e., 10, the difference in the values

\[= 8 . 794 - 8 . 750 = 0 . 044\]
∴  For the difference of (673 - 670), i.e., 3, the difference in the values 
\[= \frac{0 . 044}{10} \times 3 = 0 . 0132 = 0 . 013\]  (upto three decimal places)
∴ \[\sqrt[3]{673} = 8 . 750 + 0 . 013 = 8 . 763\]
Now \[\sqrt[3]{1346} = \sqrt[3]{2} \times \sqrt[3]{673} = 1 . 260 \times 8 . 763 = 11 . 041\] (upto three decimal places)

Thus, the answer is 11.041.

 
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Chapter 4: Cubes and Cube Roots - Exercise 4.5 [Page 36]

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R.D. Sharma Mathematics [English] Class 8
Chapter 4 Cubes and Cube Roots
Exercise 4.5 | Q 8 | Page 36

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