Question

cube root 8 + cube root 27 - cube root 64​

Answer 1

$\sqrt[3]{8} + \sqrt[3]{27} - \sqrt[3]{64} \\ \\ = 2 + 3 - 4 \\ \\ = 5 - 4 \\ \\ = 1$

$2 \times 2 \times 2 = 8 \\ 3 \times 3 \times 3 = 27 \\ 4 \times 4 \times 4 = 64$

Answer 2

$\huge\underline\mathrm\color{violet}{Answer}$

We know that ,

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

$\sqrt[3]{27} = \sqrt[3]{3 \times 3 \times 3} = 3$

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

According to the Question ,

$= > \: \sqrt[3]{8} + \sqrt[3]{27} - \sqrt[3]{64}$

$= > \: 2 + 3 - 4$

$= > \: 5 - 4$

$= > \: 1$