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\frac{1}{a^{2}-b c}+\frac{1}{b^{2}-c a}+\frac{1}{c^{2}-a b}=0 |
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Answer» 1/(a^2-bc ) + 1/(c^2-ab) + 1/(a^2-bc)Put -ac = ab + bc ; -ab = ac+ bc and -bc =ab + acyou get :=(1/(a^2+ab+ac))+(1/(c^2+ac+bc))+(1/(a^2+ab+ac))=(1/a(a+b+c))+(1/b(a+b+c))+(1/c(a+b+c))By taking LCM we get=(ab + bc + ca)/[(a+b+c)(abc)]put, ab+bc+cavalue = 0/[(a+b+c)(abc)] = 0 great |
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