Web= (2a) 3 + b 3 + 3 × 2a × b (2a + b) By further calculation = 8a 3 + b 3 + 6ab (2a + b) So we get = 8a 3 + b 3 + 12a 2 b + 6ab 2 8. (i) (3x + 1/x)3 (ii) (2x – 1)3 Solution: (i) (3x + 1/x) 3 It can be written as = (3x) 3 + (1/x) 3 + 3 × 3x × 1/x (3x + 1/x) By further calculation = 27x 3 + 1/x 3 + 9 (3x + 1/x) So we get = 27x 3 + 1/x 3 + 27x + 9/x

If a + b + 2c = 0, prove that a3 + b3 + 8c3 = 6abc.-8 - Brainly.in

Web25 mrt. 2024 · Sorted by: 1 Sonnhard got the right idea, but made one small typo. The correct answer is a 3 − b 3 + 8 c 3 + 6 a b c = ( 2 c + a − b) ( a 2 + b 2 + 4 c 2 + a b − 2 a c + 2 b c). The expression a 3 − b 3 + 8 c 3 − 6 a b c is irreducible over Z. Share Cite Follow edited Mar 25, 2024 at 14:04 answered Mar 25, 2024 at 12:53 jlammy 8,974 15 36 1 Web22 mrt. 2024 · Q.11 Factorize the following by using a suitable identity: - 16116071. A battery with 20 % 20 , percent of its full capacity is connected to a charger every minute that passes, an additional 5 % 5%5, percent of its capaci … budapest ghetto facts

Factorise the following by using a suitable identity: a. a3+b3+8c3+6abc …

Web5 jun. 2024 · => a³ + b³ + 8c³ = 6abc. Alternative Method:-We know that. If a + b + c = 0 then a³ + b³ + c³ = 3abc. We have a + b + 2c = 0. Where a = a , b = b and c = 2c. Now, If … WebIf a+b+c=0, then prove that a 3+b 3+c 3=3abc. Easy Solution Verified by Toppr We know, a 3+b 3+c 3−3abc=(a+b+c)(a 2+b 2+c 2−ab−bc−ca) Putting a+b+c=0 on RHS, we get, a 3+b 3+c 3−3abc=0 a 3+b 3+c 3=3abc, Hence proved. Was this answer helpful? 0 0 Similar questions x 3+y 3=(x+y)(x 2−xy+y 2) prove this identity. Medium View solution > WebFactorise: a3 −b3 −a+b Hard Solution Verified by Toppr We know the identity a3 −b3 = (a−b)(a2 +b2 +ab) Using the above identity, the equation a3 −b3 −a+b can be factorised as follows: a3 −b3 −a+b = (a3 −b3)−(a−b) ={(a−b)[(a)2 +(b)2 +(a×b)]}−(a−b)= [(a−b)(a2 +b2 +ab)]−(a−b) = (a−b)(a2 +b2 +ab−1) Hence, a3 −b3 −a+b =(a−b)(a2 +b2 +ab−1) budapest gatwick flights