Since (a – b), a and (a + b) are the zeroes of the polynomials \\(x^3 – 3x^2 + x + 1\\), therefore<\/p>\n
(a – b) + a + (a + b) = \\(-(-3)\\over 1\\) = 3<\/p>\n
So,\u00a0 \u00a03a = 3\u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0a = 1<\/p>\n
(a – b)a + a(a + b) + (a + b)(a – b) = \\(1\\over 1\\)<\/p>\n
\\(\\implies\\)\u00a0 \\(a^2 – ab + a^2 + ab + a^2 – b^2\\) = 1<\/p>\n
\\(\\implies\\) \\(3a^2 – b^2\\) = 1<\/p>\n
So,\u00a0 \\(3(1)^2 – b^2\\) = 1<\/p>\n
\\(\\implies\\)\u00a0 \\(b^2\\) = 2\u00a0 \u00a0or\u00a0 b = \\(\\pm \\sqrt{2}\\)<\/p>\n