TS ICET 24 May 2018 Shift 1 Paper

Question : 123 of 200

Marks: +1, -0

(3a + 2b + c)

(9a^2 + 4b^2 + c^2

- 6ab - 2bc - 3ca) =

Solution:

We know that

(x+y+z)(x^{2}+y^{2}+z^{2}-x y-y z-z x)=x^{3}+y^{3}+

z^{3}-3 x y z

∴ (3 a+2 b+c)((3 a)^{2}+(2 b)^{2}+c^{2}-(3 a)(2 b)(2 b) c-(3 c) a)

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

=27 a^{3}+8 b^{3}+c^{3}-18 a b c

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