Factorise :
9a2+19a2−2−12a+43a9a^2 + \dfrac{1}{9a^2} - 2 - 12a + \dfrac{4}{3a}9a2+9a21−2−12a+3a4
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Given,
=9a2+19a2−2−12a+43a=(3a)2+(13a)2−2−4(3a−13a)=(3a−13a)2−4(3a−13a)=(3a−13a)(3a−13a−4).\phantom{=} 9a^2 + \dfrac{1}{9a^2} - 2 - 12a + \dfrac{4}{3a} \\[1em] = (3a)^2 + \Big(\dfrac{1}{3a}\Big)^2 - 2 - 4\Big(3a - \dfrac{1}{3a}\Big) \\[1em] = \Big(3a - \dfrac{1}{3a}\Big)^2 - 4\Big(3a - \dfrac{1}{3a}\Big) \\[1em] = \Big(3a - \dfrac{1}{3a}\Big)\Big(3a - \dfrac{1}{3a} - 4\Big).=9a2+9a21−2−12a+3a4=(3a)2+(3a1)2−2−4(3a−3a1)=(3a−3a1)2−4(3a−3a1)=(3a−3a1)(3a−3a1−4).
Hence, 9a2+19a2−2−12a+43a=(3a−13a)(3a−13a−4).9a^2 + \dfrac{1}{9a^2} - 2 - 12a + \dfrac{4}{3a} = \Big(3a - \dfrac{1}{3a}\Big)\Big(3a - \dfrac{1}{3a} - 4\Big).9a2+9a21−2−12a+3a4=(3a−3a1)(3a−3a1−4).
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x2+1x2−3x^2 + \dfrac{1}{x^2} - 3x2+x21−3 in the form of factors is :
(x−1x)(x+1x)\Big(x - \dfrac{1}{x}\Big)\Big(x + \dfrac{1}{x}\Big)(x−x1)(x+x1) - 1
(x+1x+1)(x−1x−1)\Big(x + \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big)(x+x1+1)(x−x1−1)
(x−1x)(x−1x−1)\Big(x - \dfrac{1}{x}\Big)\Big(x - \dfrac{1}{x} - 1\Big)(x−x1)(x−x1−1)
(x−1x+1)(x−1x−1)\Big(x - \dfrac{1}{x} + 1\Big)\Big(x - \dfrac{1}{x} - 1\Big)(x−x1+1)(x−x1−1)
x2+14x2+1−7x−72xx^2 + \dfrac{1}{4x^2} + 1 - 7x - \dfrac{7}{2x}x2+4x21+1−7x−2x7
x2+a2+1ax+1x^2 + \dfrac{a^2 + 1}{a}x + 1x2+aa2+1x+1
x4 + y4 - 27x2y2
