We have,9×2 – 6b2x – (a4 – b4) = 0Here, Constant term = a4 – b4 = (a2 – b2)(a2\xa0+ b2)Also, the coefficient of middle term is – 6b2 = [3(a2 + b2) – 3(a2\xa0- b2)]Now using the above two values in given equation, 9×2 – 6b2x – (a4 – b4) = 0{tex}{/tex}\xa0we have, 9×2 – [3(a2 + b2) – 3(a2 – b2)]x – (a2 – b2)(a2\xa0+ b2) = 0{tex}\\Rightarrow{/tex} 9×2 – 3(a2\xa0+ b2)x + 3(a2 – b2)x – (a2 – b2)(a2 + b2) = 0{tex}\\Rightarrow{/tex} 3x[3x – (a2\xa0+ b2)] + (a2 – b2)[3x – (a2 + b2)] = 0{tex}\\Rightarrow{/tex} [3x + (a2\xa0- b2)][3x – (a2 + b2)] = 0{tex}\\Rightarrow{/tex}either [3x + (a2\xa0 – b2)] = 0 or, [3x – (a2 + b2)] = 0{tex}\\Rightarrow{/tex} 3x = -(a2 – b2) or 3x = a2 + b2{tex}\\Rightarrow x = -(\\frac{{{a^2} – {b^2}}}{3}){/tex} or {tex}x = \\frac{{{a^2} + {b^2}}}{3}{/tex}{tex}\\Rightarrow x = \\frac{{{b^2} – {a^2}}}{3}{/tex} or {tex}x = \\frac{{{a^2} + {b^2}}}{3}{/tex}Hence, the roots of given quadratic equation are\xa0{tex}\\frac{{b^2 – a^2}}{3}{/tex}and\xa0{tex}\\frac{{a^2 + b^2}}{3}{/tex}