Question

plz solve in step
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Answer 1

Okay so here in this sum we are gonna apply a special identity-
a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)
so from here we can say that,
if (a+b+c)=0,
then, a^3+b^3+c^-3abc=0
In our sum, it is given,
x=1/3(2a+3b+3c)
or, 3x-2a-3b-3c=0.................equation 1

in the Required To Prove part,
we basically have to prove,
(x-2a)^3+(x-3b)^3+(x-3c)^3-3(x-2a)(x-3b)(x-3c)=0
you can easily see that the left hand side of this equation is similar to the left hand side of the identity, a^3+b^3+c^3-3abc= (a+b+c)(a^2+b^2+c^2-ab-bc-ca)
the left hand side will be equal to zero when (a+b+c)=0.
in our sum, that means,
(x-2a)+(x-3b)+(x-3c)=0
or, 3x-2a-3b-3c=0----this has been previously proved by equation 1.
so we can say,
(x-2a)^3+(x-3b)^3+(x-3c)^3-3(x-2a)(x-3b)(x-3c)=0
or, (x-2a)^3+(x-3b)^3+(x-3c)^3= 3(x-2a)(x-3b)(x-3c)-----------[proved]

hope it helped!

Answer 2

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