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If $a^{2}+b^{2}+c^{2}=0$, then what is $\frac{\left(a^{4}-b^{4}\right)^{3}+\left(b^{4}-c^{4}\right)^{3}+\left(c^{4}-a^{4}\right)^{3}}{\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}}$
equal to ?
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equal to ?
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Verified Answer
The correct answer is:
$-a^{2} b^{2} c^{2}$
We know that if $\mathrm{x}+\mathrm{y}$ ' Hereboth in nominator and denominator $a^{4}-b^{4}+b^{4}-c^{4}+c^{4}-a^{4}=0$
and $a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}=0$
Hence, $\frac{\left(a^{4}-b^{4}\right)^{3}+\left(b^{4}-c^{4}\right)^{3}+\left(c^{4}-a^{4}\right)^{3}}{\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}}$
$=\frac{3\left(a^{4}-b^{4}\right)\left(b^{4}-c^{4}\right)\left(c^{4}-a^{4}\right)}{3\left(a^{2}-b^{2}\right)\left(b^{2}-c^{2}\right)\left(c^{2}-a^{2}\right)}$
$=\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right.$
$=\left(-c^{2}\right)\left(-a^{2}\right)\left(-b^{2}\right)$
$=-a^{2} b^{2} c^{2}$
$\left(\because a^{2}+b^{2}+c^{2}=0\right)$
and $a^{2}-b^{2}+b^{2}-c^{2}+c^{2}-a^{2}=0$
Hence, $\frac{\left(a^{4}-b^{4}\right)^{3}+\left(b^{4}-c^{4}\right)^{3}+\left(c^{4}-a^{4}\right)^{3}}{\left(a^{2}-b^{2}\right)^{3}+\left(b^{2}-c^{2}\right)^{3}+\left(c^{2}-a^{2}\right)^{3}}$
$=\frac{3\left(a^{4}-b^{4}\right)\left(b^{4}-c^{4}\right)\left(c^{4}-a^{4}\right)}{3\left(a^{2}-b^{2}\right)\left(b^{2}-c^{2}\right)\left(c^{2}-a^{2}\right)}$
$=\left(a^{2}+b^{2}\right)\left(b^{2}+c^{2}\right)\left(c^{2}+a^{2}\right.$
$=\left(-c^{2}\right)\left(-a^{2}\right)\left(-b^{2}\right)$
$=-a^{2} b^{2} c^{2}$
$\left(\because a^{2}+b^{2}+c^{2}=0\right)$
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