Question
Making use of the cube root table, find the cube root 1346
✓
Answer
By prime factorisation, we have:1346 = 2 × 673
$\Rightarrow\sqrt[3]{1346}$ $=\sqrt[3]{2}\times\sqrt[3]{673}$ Also 670 < 673 < 680 $\Rightarrow\sqrt[3]{670}<\sqrt[3]{673}<\sqrt[3]{680}$ For the difference (680 - 670), i.e., 10, the difference in the values = 8.794 - 8.750 = 0.044 $\therefore$ For the difference of (673 - 670), i.e., 3, the difference in the values $=\frac{0.004}{10}\times3= 0.013$ (upto three decimal places) $\therefore\sqrt[3]{673}$ $=8.750 +0.013$ $=8.763$ Now $\sqrt[3]{1346}$ $=\sqrt[3]{2}\times=\sqrt[3]{673}$ $=1.260\times8.763$ $=11.041$ (Up to three decimal places) Thus, the answer is 11.041.
$\Rightarrow\sqrt[3]{1346}$ $=\sqrt[3]{2}\times\sqrt[3]{673}$ Also 670 < 673 < 680 $\Rightarrow\sqrt[3]{670}<\sqrt[3]{673}<\sqrt[3]{680}$ For the difference (680 - 670), i.e., 10, the difference in the values = 8.794 - 8.750 = 0.044 $\therefore$ For the difference of (673 - 670), i.e., 3, the difference in the values $=\frac{0.004}{10}\times3= 0.013$ (upto three decimal places) $\therefore\sqrt[3]{673}$ $=8.750 +0.013$ $=8.763$ Now $\sqrt[3]{1346}$ $=\sqrt[3]{2}\times=\sqrt[3]{673}$ $=1.260\times8.763$ $=11.041$ (Up to three decimal places) Thus, the answer is 11.041.
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