Ask your question hereif x=a(b-C) y=b(c-a) z= c(a-b) then find (x^3/a^3)+(y^3/b^3)+(z^3/c^3)=
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sub. the value of x, y,z we get
=[a^3(b-c)^3]/a^3+[b^3(c-a)^3]/b^3+[c^3(a-b)^3]/c^3
=[(b-c)^3+(c-a)^3+(a-b)^3]
now
a^3+b^3=(a+b)(a^2+b^2-ab)
so
=[(b-c+c-a)((b-c)^2+(c-a)^2-(b-c)(c-a))+(a-b)^3]
=[(b-a)(b^2+c^2-2bc+a^2+c^2-2ac-bc+ab+c^2-ca)--(b-a)^3]
=[(b-a)][b^2+3c^2-3bc+a^2-3ac+ab-a^2-b^2+2ab]
=(b-a)[3ab-3bc-3ac]
=3[ab-bc-ac](b-a)
=[a^3(b-c)^3]/a^3+[b^3(c-a)^3]/b^3+[c^3(a-b)^3]/c^3
=[(b-c)^3+(c-a)^3+(a-b)^3]
now
a^3+b^3=(a+b)(a^2+b^2-ab)
so
=[(b-c+c-a)((b-c)^2+(c-a)^2-(b-c)(c-a))+(a-b)^3]
=[(b-a)(b^2+c^2-2bc+a^2+c^2-2ac-bc+ab+c^2-ca)--(b-a)^3]
=[(b-a)][b^2+3c^2-3bc+a^2-3ac+ab-a^2-b^2+2ab]
=(b-a)[3ab-3bc-3ac]
=3[ab-bc-ac](b-a)
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