Question

Factorise : i) a³-8b³-64c³-24abc .
ii) 27x³+125y³
iii) 9(2a-b)²-4(2a-b)-13
iv) 2x²+3√3 x+3 ​

Answer 1

Answer:

Step-by-step explanation:

i) (a)³-(2b)³-(4c)³-3(a)(2b)(4c)

[a³-b³-c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)

=>(a)³-(2b)³-(4c)³-3(a)(2b)(4c)=(a+2b+4c)[(a)²+(2b)²+(4c)²-(a)(2b)-(2b)(4c)-(4c)(a)

=>(a+2b+4c)(a²+4b²+16c²-2ab-8bc-4ca)

ii) 27x3+125y3

(3x)3+(5y)3

according to a3 +b3 identity

(a+b)(a2+b2-ab)

(3x+5y)(9x2+25y2-15xy)

iii) Given 9(2a - b)^2 - 4(2a - b) - 13

= > 9(2a - b)(2a - b) + 9(2a - b) - 13(2a - b) - 13

= > 9(2a - b)[(2a - b) + 1] - 13[(2a - b) + 1]

= > (2a - b + 1)(18a - 9b - 13).

iv ) In attachment

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