Solve the following quadratic equations by factorization:

$9 x^{2} - 6 b^{2} x - (a^{4} - b^{4}) = 0$

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Solution

$9 x^{2} - 6 b^{2} x - (a^{4} - b^{4}) = 0 \Rightarrow 9 x^{2} - 6 b^{2} x - (a^{2} - b^{2}) (a^{2} + b^{2}) = 0 \Rightarrow 9 x^{2} + 3 (a^{2} - b^{2}) x - 3 (a^{2} + b^{2}) x - (a^{2} - b^{2}) (a^{2} + b^{2}) = 0 \Rightarrow 3 x [3 x + (a^{2} - b^{2})] - (a^{2} + b^{2}) [3 x + (a^{2} - b^{2})] = 0 \Rightarrow [3 x - (a^{2} + b^{2})] [3 x + (a^{2} - b^{2})] = 0 \Rightarrow 3 x - (a^{2} + b^{2}) = 0 or 3 x + (a^{2} - b^{2}) = 0 \Rightarrow x = \frac{a^{2} + b^{2}}{3} or x = - \frac{a^{2} - b^{2}}{3} \Rightarrow x = \frac{a^{2} + b^{2}}{3} or x = \frac{b^{2} - a^{2}}{3}$

Hence, the factors are $\frac{a^{2} + b^{2}}{3}$ and $\frac{b^{2} - a^{2}}{3}$ .

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