Question

Q21. Simplify:
(4x2 – 9y2)3 + (9y2 – 16y2)3 + (1632 – 4x2)3
(2x – 3y)3
+ (3y – 4z ) + (4z – 2x ) 3​

Answer 1

We know the corollary: if a+b+c=0 then a

3

+b

3

+c

3

=3abc

Using the above corollary taking a=(2x−3y), b=(4z−2x) and c=(3y−4z), we have a+b+c=2x−3y+4z−2x+3y−4z=0 then the value of (2x−3y)

3

+(4z−2x)

3

+(3y−4z)

3

is:

(2x−3y)

3

+(4z−2x)

3

+(3y−4z)

3

=3[(2x−3y)×(4z−2x)×(3y−4z)]=3(2x−3y)(4z−2x)(3y−4z)

=6(2x−3y)(2z−x)(3y−4z)

Hence, (2x−3y)

3

+(4z−2x)

3

+(3y−4z)

3

=6(2x−3y)(2z−x)(3y−4z)

hope it is helpfull