1. If $$a^{2}+b^{2}+c^{2}+\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=6$$, then what is the value of $$(a^{2}+b^{2}+c^{2})$$ ?
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By: anil on 05 May 2019 03.04 pm
Given : $$a^{2}+b^{2}+c^{2}+frac{1}{a^{2}}+frac{1}{b^{2}}+frac{1}{c^{2}}=6$$ => $$(a^2+frac{1}{a^2}-2)+(b^2+frac{1}{b^2}-2)+(c^2+frac{1}{c^2}-2)=0$$ => $$(a-frac{1}{a})^2+(b-frac{1}{b})^2+(c-frac{1}{c})^2=0$$ $$ecause$$ Sum of three positive terms is zero, hence each term is equal to 0. => $$(a-frac{1}{a})=$$ $$(b-frac{1}{b})=$$ $$(c-frac{1}{c})=0$$ => $$frac{a^2-1}{a}=0$$ => $$a^2=1$$ Similarly, $$b^2=c^2=1$$ $$ herefore$$ $$(a^{2}+b^{2}+c^{2})=1+1+1=3$$ => Ans - (A)
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