Question
$$(x-4)^2(x+4)-(x-4)(x+4)^2+3(x^2-16) = 0$$
Answer
$$ \begin{matrix}x_1 = 4 & x_2 = -4 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} (x-4)^2(x+4)-(x-4)(x+4)^2+3(x^2-16) &= 0&& \text{simplify left side} \\[1 em](x^2-8x+16)(x+4)-(x-4)(x^2+8x+16)+3(x^2-16) &= 0&& \\[1 em]x^3+4x^2-8x^2-32x+16x+64-(x^3+8x^2+16x-4x^2-32x-64)+3x^2-48 &= 0&& \\[1 em]x^3-4x^2-16x+64-(x^3+4x^2-16x-64)+3x^2-48 &= 0&& \\[1 em]x^3-4x^2-16x+64-x^3-4x^2+16x+64+3x^2-48 &= 0&& \\[1 em]-8x^2+128+3x^2-48 &= 0&& \\[1 em]-5x^2+80 &= 0&& \\[1 em] \end{aligned} $$
$ -5x^{2}+80 = 0 $ is a quadratic equation.
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