Question

a=2x-1
b=2x-2
c=3-4x
Then,a³+b³+c³=?
​

Answer 1

• Step-by-step explanation:
• $4x + (2x - 1)(2x + 1)$

Answer 2

Answer:

Here we will use an identity

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

Here given

a= 2x-1

b = 2x-2

c = 3-4x

a+b+c = 2x-1+2x-2+3-4x

=2x+2x-4x-1-2+3

=4x-4x-3+3

=0

So

$a {}^{3} + b {}^{3} + c {}^{3} - 3 abc = 0 \times( a {}^{2} + b {}^{2} + c {}^{2} - ab - bc - ca)$

$a {}^{3} + b {}^{3} + c {}^{3} - 3 abc = 0 \times( a {}^{2} + b {}^{2} + c {}^{2} - ab - bc - ca) \\ a {}^{3} + b {}^{3} + c {}^{3} - 3abc = 0 \\ a {}^{3} + b {}^{3} + c {}^{3} = 3 abc \\a fter \: putting \: the \: value \: of \: a \: \: \: b \: \: \: and \: \: \: c \\ we \: get \: a {}^{3} + b {}^{3} + c {}^{3} = 3 \times (2x - 1)(2x - 2)(3 - 4x) \\ = 3 \times (2x - 1) \times ( - 8x {}^{2} + 14x - 6) \\ 3 \times ( - 16x {}^{3} + 36x {}^{2} - 26x + 6 \\ = - 48x {}^{3} + 108x {}^{2} - 78x + 18$