`2a-3b-[4a-3b-{a-2c-(a-2b-c)}]`

Prove that a^3-(2a-b-c)^3+(b-2c)^3-(2b-c-a)^3=3(b+c-a)(a+b-2c)(2a-2b-c)

If (5a-3b)/a= (4a+b-2c)/(a+4b-2c) = (a+2b-3c)/(4a-4c) , then show that 6a=4b=3c.

if (2a+3b+4c+5d)/(2a-3b-4c-5d)=(2a-3b+4c-5d)/(2a-3b-4c+5d) , then show that a,3b,2c and 5d are in proportion.

Simplify: 2a - [3b - {a - (2c - 3b) + 4c - 3 (a - b - 2c)}]

Subtract the sum of 2a+3b,a-2b+c,-a+2c and 4a+2b-5 from 6a-4b+8

Simplify : (Use formulae) (2a-3b+4c)^2-(2a+3b-4c)^2

Simplify : ( 2a - 3b + 3c ) ( 4a + 2b - c )

Simplify the following : (a+2b+3c)(a+b)-(a-b-c)(2a+3c)+(2a+b)(2c-2a-4b)

Simplify (a+b)(2a-3b+c)-(2a-3b)c

Simplify (a+b)(2a-3b+c)-(2a-3b)c.