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If $a+b+c=0$ and $a^2+b^2+c^2=36$, what is the value of $a^2b^2+b^2c^2+c^2a^2$?

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2 Answers 2

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We'll work with what we have until we get what we want:

$$0 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2 ac = 36 + 2(ab + bc + ac)$$ And therefore we can conclude that $$ - 18 = ab + bc + ac $$ This in turn means that $$ 324 = (ab + bc + ac) ^2 = a^2b^2 + b^2 c^2 + a^2 c^2 + 2(a^2bc + ab^2 c + abc^2) \\\\ =a^2b^2 + b^2 c^2 + a^2 c^2 + 2abc(a + b + c) = a^2b^2 + b^2 c^2 + a^2 c^2 + 0$$

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  • $\begingroup$ @labbhattacharjee I haven't, but it's multiplied with $(a + b + c) = 0$ $\endgroup$
    – Arthur
    Jul 29, 2013 at 11:35
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$$(a+b+c)=0\implies (a+b+c)^2=0 $$ $$\implies a^2+b^2+c^2+2(ab+bc+ca)=0$$ $$\implies 36+2(ab+bc+ca)=0$$ $$\implies (ab+bc+ca)=-18$$ $$\implies (ab+bc+ca)^2=324$$

just solve this whole square you can find answer

$$\implies a^2b^2+b^2c^2+c^2a^2+2(ab^2c+bc^2a+a^2bc)=324$$ $$\implies a^2b^2+b^2c^2+c^2a^2+2abc(b+c+a)=324$$ since $a+b+c=0$

so $$\implies a^2b^2+b^2c^2+c^2a^2=324$$

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