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    • Important Maths Formulas For Competitive Exams - Assam Job Alert



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    Important Maths Formulas For Competitive Exams

    Important Maths Formulas For Competitive Exams of SSC, FCI, IBPS, SBI, APDCL etc.



    Important Maths Formulas For you upcoming Competitive Exams – As competitive exams are frequent in today’s date therefore, it is important get updated for upcoming challenges. Though the shortcut methods for aptitude paper are in trend for the competitive exams but it very important to have knowledge about the basic formulas/process of solving the maths problems in conventional way.

    Gauhati High Court Recruitment 2019 For Judicial Assistant

    Following formulas are well known for you as you had solve maths problems in your High School level using those formula. Which are easy to remember as you have already memorised in your school days. In past few months there are large vacancies are published by Railway Recruitment Board (RRB), Food Corporation of India (FCI), Staff Selection Commission (SSC), Institute of Banking Personnel Selection (IBPS), State Bank of India (SBI), Allahbad Bank etc organisations. In the recruitment process the organisation will conduct computer based written test and to solve aptitude question it important to have knowledge of the following formulas.



    Important Maths Formulas For Competitive Exams:

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    • (a + b) 2= a2 + 2ab + b2
    • (a+b)2=(a-b)2+4ab
    • (a – b) 2= a2 – 2ab + b2
    • (a-b)2=(a+b)2-4ab
    • a2 + b2 = (a+b)2-2ab 
    • a2+b2=(a-b)2+2ab
    • a2 – b2 = (a + b) – (a – b)
    • 2 (a2+b2) = (a + b) 2 + (a – b) 2
    • (a+b+c)2=(a2+b2+c2)+2(ab+bc+ca)
    • (a2 + b2 + c2) = (a + b + c) 2 – 2(ab + be + ca)
    • 2 (ab + bc+ca) = (a + b+c)2-(a2+b2+c2)
    • (a + b) 3= a3 + 3a2b + 3ab2 + b3
    • (a+b)3=a3+b3+3ab(a+b)
    • (a – b) 3= a3 – 3a2b + 3ab2 – b3
    • (a – b) 3= a3 – b3 – 3ab (a – b)
    • a3 + b3 = (a + b) (a2 – ab + b2)
    • a3 + b3 = (a + b) 3 – 3ab (a + b)
    • a3 – b3 = (a – b) (a2 + ab + b2)
    • a3 – b3 = (a – b) 3+ 3ab (a – b)
    • (a+b+c)3=a3+b3+c3+3 (a+b)(b+c)(c+a)
    • a3 + b3 + c3– 3abc = (a + b + c) (a2+ b2 + c2 – ab – bc – ca)
    • a3+b3+c3-3abc=1/2(a+ b+c) {(a-b)2+(b-c)2+(c-a)2}
    • (x+a)(x+b)=x2+(a+b)x+ab
    • (x+a)(x-b)=x2+(a-b)x-ab
    • (x-a)(x+b)=x2+(b-a)x-ab
    • (x-a)(x-b) =x2-(a + b) x + ab
    • (x+p)(x+q)(x+r)=x3+(p+q+r)x2+(pq+qr+rp)x+pqr
    • bc (b – c) + ca (c – a) + ab (a – b) = – (b – c) (c – a) (a – b)
    • a2 (b – c) + b2 (c – a) + c2 (a – b) = – (b – c) (c – a) (a – b)
    • a (b2 – c2) + b (c2 – a2) + c (a2 – b2) = (b – c) (c – a) (a – b)
    • a3 (b – c) + b3 (c – a) + c3 (a – b) = – (b – c) (c – a) (a – b) (a + b + c)
    • b2c2(b2-c2) + c2a2(c2-a2) + a2b2(a2-b2) = – (b-c) (c-a) (a-b) (b+c) (c+a) (a+b)
    • (ab+bc+ca)(a+b+c)-abc=(a+b)(b+c)(c+a)
    • (b+c)(c+a)(a+b)+abc=(a+b+c)(ab+bc+ca)



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