Question

Solve for the value of x:
$9x^2-6b^{2}x-(a^4-b^4)=0$

Answer 1

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

$9x^2-3\left({2b}^2\right)x-(a^4-b^4)=0$

$9x^2-3(b^2-a^2+b^2+a^2)x+(b^4-a^4)=0$

$3x(3x-b^2+a^2)-(b^2+a^2)(3x-b^2+a^2)=0$

$(3x-b^2-a^2)(3x-b^2+a^2)=0$

Now, $3x-b^2-a^2=03x=a^2+b^{2}x=\frac{a^2+b^2}{3}$ or $3x-b^2+a^2=03x=b^2-a^{2}x=\frac{b^2-a^2}{3}$

So, $x=\frac{a^2+b^2}{3},\frac{b^2-a^2}{3}$