If a+b+c=0, then find the value of frac{a^{4} | vedclass

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(C) Given $a+b+c=0.$Squaring both sides,we get $(a+b+c)^2 = 0^2.$$a^2+b^2+c^2+2(ab+bc+ca) = 0 \Rightarrow a^2+b^2+c^2 = -2(ab+bc+ca).$Squaring again,$

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$2$

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(C) Given $a+b+c=0.$Squaring both sides,we get $(a+b+c)^2 = 0^2.$$a^2+b^2+c^2+2(ab+bc+ca) = 0 \Rightarrow a^2+b^2+c^2 = -2(ab+bc+ca).$Squaring again,$(a^2+b^2+c^2)^2 = [-2(ab+bc+ca)]^2.$$a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2) = 4(a^2b^2+b^2c^2+c^2a^2+2abc(a+b+c)).$Since $a+b+c=0,$ the term $2abc(a+b+c) = 0.$So,$a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2) = 4(a^2b^2+b^2c^2+c^2a^2).$$a^4+b^4+c^4 = 4(a^2b^2+b^2c^2+c^2a^2) - 2(a^2b^2+b^2c^2+c^2a^2) = 2(a^2b^2+b^2c^2+c^2a^2).$Therefore,$\frac{a^4+b^4+c^4}{a^2b^2+b^2c^2+c^2a^2} = 2.$

If $a+b+c=0,$ then find the value of $\frac{a^{4}+b^{4}+c^{4}}{b^{2} c^{2}+c^{2} a^{2}+a^{2} b^{2}}.$

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