Find the cubes of:
(i) −11
(ii) −12
(iii) −21

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Solution

(i)
Cube of $-$ 11 is given as:
${(- 11)}^{3} = - 11 \times - 11 \times - 11 = - 1331$
Thus, the cube of 11 is ( $-$ 1331).

(ii)
Cube of $-$ 12 is given as:
${(- 12)}^{3} = - 12 \times - 12 \times - 12 = - 1728$

Thus, the cube of $-$ 12 is ( $-$ 1728).

(iii)
Cube of $-$ 21 is given as:
${(- 21)}^{3} = - 21 \times - 21 \times - 21 = - 9261$

Thus, the cube of $-$ 21 is ( $-$ 9261).

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PLS show with working.......

1) Find the cube root of:

i) 0.000125

\frac{x^{2} + 1}{x} = 2 \frac{1}{2}

x - \frac{1}{x}

x^{3} - \frac{1}{x^{3}}

\frac{7}{9}

- \frac{8}{11}

\frac{12}{7}

- \frac{13}{8}

2 \frac{2}{5}

3 \frac{1}{4}

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