Without actually calculating cubes, value of each of following (21)^{3}+(15)^{3}+(-36)^{3}

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2. Polynomials

hard

Without actually calculating the cubes, find the value of each of the following

$(21)^{3}+(15)^{3}+(-36)^{3}$

A

$61280$

B

$-34020$

C

$65041$

D

$-53120$

Solution

Taking $a=21 ; \quad b=15 ; c=(-36),$ we get

$a+b+c=21+15+(-36)$

$=36-36=0$

Now, if $a+b+c=0,$ then

$a^{3}+b^{3}+c^{3}=3 a b c$

$\therefore(21)^{3}+(15)^{3}+(-36)^{3}=3(21)(15)(-36)$

$=63 \times(-540)$

$=-34020$

Standard 9

Mathematics

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