# Algebra Questions In Hindi PDF For All Competitive Exams

**Algebra Questions In Hindi PDF**:- Today, we are sharing the** Maths Algebra Question ** for you. This Algebra Questions 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 **M****aths Algebra Question in hindi** 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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