Question

$\sqrt3{648}+\sqrt3{2000}$

Explanation:

Step1: Simplify $\sqrt[3]{648}$

Factor 648: $648 = 8\times81 = 8\times9\times9 = 2^3\times3^4 = 2^3\times3^3\times3$. So $\sqrt[3]{648}=\sqrt[3]{2^3\times3^3\times3}=2\times3\times\sqrt[3]{3}=6\sqrt[3]{3}$.

Step2: Simplify $\sqrt[3]{2000}$

Factor 2000: $2000 = 1000\times2 = 10^3\times2$. So $\sqrt[3]{2000}=\sqrt[3]{10^3\times2}=10\sqrt[3]{2}$. Wait, no, wait, 2000 is $10^3\times2$? Wait, 10^3 is 1000, 1000×2=2000, yes. But wait, maybe I made a mistake in 648. Wait, 648 divided by 8 is 81, 81 is 9×9, 9 is 3², so 648=8×81=2³×3⁴=2³×3³×3, so cube root of 2³×3³×3 is 2×3×cube root of 3=6∛3. And 2000=1000×2=10³×2, so cube root of 10³×2 is 10∛2? Wait, no, wait, maybe we can calculate the numerical values. Let's calculate the approximate values. $\sqrt[3]{648}\approx8.653$, $\sqrt[3]{2000}\approx12.6$. Then add them: 8.653 + 12.6 = 21.253. Wait, but maybe we can simplify further? Wait, no, maybe the problem is to calculate the sum of the cube roots. Let's use a calculator approach. $\sqrt[3]{648}\approx8.65255$, $\sqrt[3]{2000}\approx12.59921$. Then sum: 8.65255 + 12.59921 = 21.25176.

Answer:

Approximately 21.25 (or more precisely, $6\sqrt[3]{3}+10\sqrt[3]{2}$ or the decimal approximation as above).