Question

which two integers is $sqrt3{96}$ between? 19 and 20 4 and 5 7 and 8 9 and 10

Explanation:

Step1: Find cube of 4

$4^3 = 4 \times 4 \times 4 = 64$

Step2: Find cube of 5

$5^3 = 5 \times 5 \times 5 = 125$

Step3: Compare values

$64 < 96 < 125$, so $\sqrt[3]{64} < \sqrt[3]{96} < \sqrt[3]{125}$, which simplifies to $4 < \sqrt[3]{96} < 5$

Answer:

4 and 5