
Algebra Questions With Solution
Algebra Maths Formulas And Notes PDF:- Today, we are sharing the Algebra Question for you. This Algebra Questions with Solution PDF can prove to be important for the preparation of upcoming government exams like SSC CGL, BANK, RAILWAYS, RRB NTPC, LIC AAO, and many other exams. Maths trick question notes are very important for any Sarkari exam. This Maths PDF is being provided to you free of charge, which you can DOWNLOAD by clicking on the DOWNLOAD button given below, and you can also DOWNLOAD some more new PDFs related to this Maths tricks competitive exams by going to the related notes. You can learn about all the new updates on PDFDOWNLOAD.IN by clicking on the Allow button on the screen.
pdfdownload.in is an online educational website, where we are sharing Algebra Question PDF for free DOWNLOAD for UPSC, SSC, BANK, RAILWAY, LIC, and many other exams. Our General Science Questions PDF is very simple and easy to understand. We also cover basic subjects like Mathematics, Geography, History, General Science, Politics, etc. We also share study material including previous year question papers, current affairs, important sources, etc. for upcoming government exams. Our PDF will help you prepare for any SARKARI EXAM.
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Important Maths Formula
- प्राकृतिक संख्या (Natural Numbers) – an – bn = (a – b)(an-1 + an-2 +…+ bn-2a + bn-1)
- सम संख्या (Even) – (n = 2k), an + bn = (a + b)(an-1 – an-2b +…+ bn-2a – bn-1)
- विषम संख्या (Odd) – (n = 2k + 1), an + bn = (a + b)(an-1 – an-2b +…- bn-2a + bn-1)
- (a + b + c + …)2 = a2 + b2 + c2 + … + 2(ab + ac + bc + ….
- घातांक के नियम (Low Of Formula Exponents)
- 1. (am)(an) = am+n
- 2. (ab)m = ambm
- 3. (am)n = amn
- a2 – b2 = (a – b)(a + b)
- (a+b)2 = a2 + 2ab + b2
- a2 + b2 = (a – b)2 + 2ab
- (a – b)2 = a2 – 2ab + b2
- (a + b + c)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc
- (a – b – c)2 = a2 + b2 + c2 – 2ab – 2ac + 2bc
- (a + b)3 = a3 + 3a2b + 3ab2 + b3 ; (a + b)3 = a3 + b3 + 3ab(a + b)
- (a – b)3 = a3 – 3a2b + 3ab2 – b3
- a3 – b3 = (a – b)(a2 + ab + b2)
- a3 + b3 = (a + b)(a2 – ab + b2)
- (a + b)3 = a3 + 3a2b + 3ab2 + b3
- (a – b)3 = a3 – 3a2b + 3ab2 – b3
- (a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4)
- (a – b)4 = a4 – 4a3b + 6a2b2 – 4ab3 + b4)
- a4 – b4 = (a – b)(a + b)(a2 + b2)
- a5 – b5 = (a – b)(a4 + a3b + a2b2 + ab3 + b4)
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