find the vale of x3-2x2-7x+5
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x^3–2x^2–7x+5 when x=1/2+Г3. (Г stands for square root)
=(1/2+Г3)^3–2(1/2+Г3)^2–7(1/2+Г3)+5
=1–2(2+Г3)-7(2+Г3)^2+5(2+Г3)^3 /(2+Г3)^3
(taking Lcm….here highest power of (2+Г3)
=1-4-2Г3–7(7+4Г3)+5(26+15Г3) /(26+15Г3)
= 78+45Г3 / 26+15Г3
=3(26+15Г3) / (26+15 Г3)
=3
=(1/2+Г3)^3–2(1/2+Г3)^2–7(1/2+Г3)+5
=1–2(2+Г3)-7(2+Г3)^2+5(2+Г3)^3 /(2+Г3)^3
(taking Lcm….here highest power of (2+Г3)
=1-4-2Г3–7(7+4Г3)+5(26+15Г3) /(26+15Г3)
= 78+45Г3 / 26+15Г3
=3(26+15Г3) / (26+15 Г3)
=3
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find all angles of rhombus if adjacent angle are in the ratio 3:5
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