Question

prove that (2a+3b-4c)3-8a3-27b3+64c3=6(2a+3b)(3b-4c)(a-2c)

Answer 1

Answer:

We have,

2a+3b+c=0

2a+3b=−c …….. (1)

On taking cube both sides, we get

(2a+3b)

3

=(−c)

3

8a

3

+27b

3

+3×2a×3b(2a+3b)=−c

3

8a

3

+27b

3

+18ab(2a+3b)=−c

3

From equation (1),

8a

3

+27b

3

+18ab(−c)=−c

3

8a

3

+27b

3

−18abc=−c

3

8a

3

+27b

3

+c

3

=18abc

Hence, proved.