$\frac{(20^{3} + 40^{3})}{60} =$ ______________.

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Solution

$a^{3} + b^{3} = (a^{2} + b^{2} - a b) (a + b)$

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Similar questions

Express each of the of the following rations in its simplest from:

(i) 60:72
(ii) 324:144
(iii) 85:391
(iv) 186:403

i^{403}

tan 203^{0} + tan 22^{0} + tan 203^{0} tan 22^{0} =

(2198 - 1347 - 403) \div (159 - 113 - 27) = ?

\int \frac{c o {sec}^{2} x - 203}{{(cos x)}^{203}} d x

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