Factorise the following:
x2+1x2−11x^2 + \dfrac{1}{x^2} - 11x2+x21−11.
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x2+1x2−11=x2+1x2−2−9=(x2+1x2−2)−(3)2=(x2−2×x2×1x2+1x2)−32x^2 + \dfrac{1}{x^2} - 11 = x^2 + \dfrac{1}{x^2} - 2 - 9 \\[1em] = \Big(x^2 + \dfrac{1}{x^2} - 2\Big) - (3)^2 \\[1em] = \Big(x^2 - 2 \times x^2 \times \dfrac{1}{x^2} + \dfrac{1}{x^2}\Big) - 3^2x2+x21−11=x2+x21−2−9=(x2+x21−2)−(3)2=(x2−2×x2×x21+x21)−32
We know that,
(a - b)2 = a2 - 2ab + b2
and
a2 - b2 = (a - b)(a + b)
(x2−2×x2×1x2+1x2)−32=(x−1x)2−32=(x−1x+3)(x−1x−3).\Big(x^2 - 2 \times x^2 \times \dfrac{1}{x^2} + \dfrac{1}{x^2}\Big) - 3^2 = \Big(x -\dfrac{1}{x}\Big)^2 - 3^2 \\[1em] = \Big(x - \dfrac{1}{x} + 3\Big)\Big(x - \dfrac{1}{x} - 3\Big).(x2−2×x2×x21+x21)−32=(x−x1)2−32=(x−x1+3)(x−x1−3).
Hence, x2+1x2−11=(x−1x+3)(x−1x−3)x^2 + \dfrac{1}{x^2} - 11 = \Big(x - \dfrac{1}{x} + 3\Big)\Big(x - \dfrac{1}{x} - 3\Big)x2+x21−11=(x−x1+3)(x−x1−3).
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(a2 - b2)(c2 - d2) - 4abcd
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x4 + 5x2 + 9
(i) x2 + 4x2\dfrac{4}{x^2}x24 - 5
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