Subjective Type

If $$a + b + c = 0$$, prove that identities $$(b^{2} c + c^{2}a + a^{2}b - 3abc)(bc^{2} + ca^{2} + ab^{2} - 3abc) = (bc + ca + ab)^{3} + 27a^{2} b^{2} c^{2}$$.

Solution

⇒(b2c+c2a+a2b−3abc)(bc2+ca2+ab2−3abc)⇒(b2c+c2a+a2b−3abc)(bc2+ca2+ab2−3abc)
=(b2c+c2a+a2b)(bc2+ca2+ab2)−3abc∑a2b+9a2b2c2⟶1=(b2c+c2a+a2b)(bc2+ca2+ab2)−3abc∑a2b+9a2b2c2⟶1

Now, (b2c+c2a+a2b)(bc2+ca2+ab2)=b3c3+c3a3+a3b3+abc(a3+b3+c3)+3a2b2c2(b2c+c2a+a2b)(bc2+ca2+ab2)=b3c3+c3a3+a3b3+abc(a3+b3+c3)+3a2b2c2
={(bc+ca+ab)3−3abc∑a2b−6a2b2c2}+abc(a3+b3+c3)+3a2b2c2={(bc+ca+ab)3−3abc∑a2b−6a2b2c2}+abc(a3+b3+c3)+3a2b2c2

Hence from 11 the given expression,
=(bc+ca+ab)3−6abc∑a2b+6a2b2c2+abc(a3+b3+c3)=(bc+ca+ab)3−6abc∑a2b+6a2b2c2+abc(a3+b3+c3)

But ∑a2b=(a+b+c)(a2+b2+c2)−3abc∑a2b=(a+b+c)(a2+b2+c2)−3abc
=−3abc∵a+b+c=0=−3abc∵a+b+c=0
Also, a3+b3+c3=3abca3+b3+c3=3abc

Hence the expression,
=(bc+ca+ab)3+18a2b2c2+6a2b2c2+3a2b2c2=(bc+ca+ab)3+18a2b2c2+6a2b2c2+3a2b2c2
=(bc+ca+ab)3+27a2b2c2=(bc+ca+ab)3+27a2b2c2

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