Find the cubes of:

(i) 2.1 (ii) 0.4 (iii) 1.6

(iv) 2.5 (v) 0.12 (vi) 0.02

(vii) 0.8

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Solution

$(i) 2.1 = (2.1)^{3} = {(\frac{21}{10})}^{3} = \frac{21 \times 21 \times 21}{10 \times 10 \times 10} = \frac{9261}{1000} = 9.261 (i i) 0.4 = (0.4)^{3} = {(\frac{4}{10})}^{3} = \frac{4 \times 4 \times 4}{10 \times 10 \times 10} = \frac{64}{1000} = 0.064 (i i i) 1.6 = (1.6)^{3} = {(\frac{16}{10})}^{3} = \frac{16 \times 16 \times 16}{10 \times 10 \times 10} = \frac{4096}{1000} = 4.096 (i v) 2.5 = (2.5)^{3} = {(\frac{25}{10})}^{3} = \frac{25 \times 25 \times 25}{10 \times 10 \times 10} = \frac{15625}{1000} = 15.625 (v) 0.12 = (0.12)^{3} = {(\frac{12}{100})}^{3} = \frac{12 \times 12 \times 12}{100 \times 100 \times 100} = \frac{1728}{1000000} = 0.001728 (v i) 0.02 = (0.02)^{3} = {(\frac{2}{100})}^{3} = \frac{2 \times 2 \times 2}{100 \times 100 \times 100} = \frac{8}{1000000} = 0.000008 (v i i) 0.8 = (0.8)^{3} = {(\frac{8}{100})}^{3} = \frac{8 \times 8 \times 8}{10 \times 10 \times 10} = \frac{512}{1000} = 0.512$

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