Complete 39 Algebra Forumula and Tricks for UPSC CSAT 2024 and 2025
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Complete 39 Algebra Forumula and Tricks for UPSC CSAT 2024 and 2025

On this page, we have compiled the complete list of short tricks and 39 formula for CSAT questions related to Algebra.

Most Important Algebra Formulae

Condition: a,b,c belong to N

  1. $$(a + b)^2 = (a + b)(a - b) = a^2 + 2ab + b^2 = (a - b)^2 + 4ab$$

  2. $$(a - b)^2 = a^2 - 2ab + b^2 = (a + b)^2 - 4ab$$

  3. $$(a^2 - b^2) = (a + b)(a - b)$$

  4. $$(a^2 + b^2) = (a + b)^2 - 2ab = (a - b)^2 + 2ab$$

  5. $$(a + b +c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = a^2 + b^2 + c^2 + 2(ab + bc + ca)$$

  6. $$(ab) = \left(\frac{a + b}{2}\right)^2 - \left(\frac{a - b}{2}\right)^2$$

  7. $$(a + b)^3 = a^3 + b^3 + 3a^2b = a^3 + b^3 + 3ab(a + b)$$

  8. $$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3 = a^3 - b^3 - 3ab(a - b)$$

  9. $$a^3 + b^3 = (a + b)(a^2 + b^2 - ab) = (a + b)^3 - 3ab(a + b)$$

  10. $$a ^ 3 - b ^ 3 = (a - b)(a ^ 2 + ab + b ^ 2) = (a - b) ^ 3 + 3ab(a - b)$$

  11. $$a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab -bc -ca)$$

  12. $$(a - b) ^ 2 = (b - a) ^ 2$$

  13. $$(a - b) + (b - c) + (c - a) = 0$$

  14. $$a(b - c) + b(c - a) + c(a - b) = 0$$

  15. $$(a - k)(b - c) + (b - k)(c - a) + (c - k)(a - b) = 0$$

  16. $$(x + a)(x + b)(x + c) = x ^ 3 + (a + b + c) x ^ 2 + (ab + bc + ca) x + abc$$

  17. $$(a + b + c) ^ 2 = a ^ 2 + b ^ 2 + c ^ 2 + 2(ab + bc + ca)$$ $$a ^ 2 + b ^ 2 + c ^ 2 = (a + b + c) ^ 2 - 2(ab + bc + ca)$$

  18. $$(a + b + c) ^ 3 = a ^ 3 + b ^ 3 + c ^ 3 + 3(a + b)(b + c)(c + a)$$ $$a ^ 3 + b ^ 3 + c ^ 3 = (a + b + c) ^ 3 - 3(a + b)(b + c)(c + a)$$

  19. $$a ^ 2(b - c) + b ^ 2(c - a) + c ^ 2(a - b) = - (a - b)(b - c)(c - a)$$

  20. $$a ^ 2 + b ^ 2 + c ^ 2 - ab - bc - ca = {1 \over 2} [(a - b) ^ 2 + (b - c) ^ 2 + (c - a) ^ 2]$$

Factorization Formulae

  1. $$x ^ 2 + (a + b) x + ab = (x + a)(x + b)$$

  2. $$x ^ 2 - (a + b) x + ab = (x - a)(x - b)$$

  3. $$a^3 + b^3 + c^3 - 3abc = {1 \over 2}(a+b+c) [(a-b)^ 2 + (b−c)^ 2 + (c-a)^ 2]$$

  4. $$a^3 - b^3 - c^3 - 3abc = (a-b-c)(a^2 + b^2+c^2 + ab - bc + ca)$$

  5. $$2b ^ 2 c ^ 2 + 2c ^ 2 c ^ 2 + 2a ^ 2 b ^ 2 - a ^ 4 - b ^ 4 - c ^ 4 = (a + b + c)(b + c - a)(c + a - b)(a + b - c)$$

  6. $$a ^ 2 (b - c) + b ^ 2 (c - a) + c ^ 2 (a - b) = - (b - c) (c - a)(a - b)$$

  7. $$bc(b - c) + ca(c - a) + ab(a - b) = - (b - c) (c - a)(a - b)$$

  8. $$a(b ^ 2 - c ^ 2) + b(c ^ 2 - a ^ 2) + c(a ^ 2 - b ^ 2) = (b - c)(c - a)(a - b)$$

  9. $$(a + b + c) ^ 3 - a ^ 3 - b ^ 3 - c ^ 3 = 3(b + c)(c + a)(a + b)$$

  10. $$4ab = (a + b) ^ 2 - (a - b) ^ 2$$

Fundamental Law of Indices

If m & n are positive natural numbers, following formulae are applicable:

  1. $$a^m \times b^n = a^{m+n}$$

  2. $${a^m \over b^n} = a^{m-n}$$

  3. $${\left(a \over b\right)}^m = {a^m \over b^m}$$

  4. $${\left(a^m\right)}^n = a^{mn}$$

  5. $${a^m \over a^n} = a^{m-n} = {1 \over a^{n-m}}$$

  6. $$a^{m \over n} = \sqrt[n]{a^m}$$

  7. $$a^{m-m} = {a^m \over a^m} = 1 [\therefore a^0 = 1]$$

  8. $${a^{-m}} = {1 \over a^m}$$

  9. $$(ab)^m = a^m \times b^m$$

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