Ta có :

a.  $(a + b) ^{2}– (a – b) ^{2}$

=  $(a ^{2}+ 2ab + b ^{2}) – (a ^{2}– 2ab + b ^{2})$

=  $a ^{2}+ 2ab + b ^{2}– a ^{2}+ 2ab - b ^{2}= 4ab$

Vậy $(a + b) ^{2}– (a – b) ^{2}=4ab$

b.  $(a + b) ^{3}– (a – b) ^{3}– 2b ^{3}$

=  $(a ^{3}+ 3a ^{2}b + 3ab ^{2}+ b^{3} ) – (a ^{3}– 3a ^{2}b + 3ab ^{2}– b ^{3}) – 2b ^{3}$

=  $a ^{3}+ 3a ^{2}b + 3ab ^{2}+ b ^{3}– a ^{3}+ 3a ^{2}b - 3ab ^{2}+ b ^{3}– 2b ^{3}=6a^{2}b$

Vậy $(a + b) ^{3}– (a – b) ^{3}– 2b ^{3}=6a^{2}b$

c.  $(x + y + z) ^{2}– 2(x + y + z)(x + y) + (x + y) ^{2}$

=  $x ^{2}+ y ^{2}+ z ^{2}+ 2xy + 2yz + 2xz – 2(x ^{2}+ xy + yx + y ^{2}+ zx + zy) + x ^{2}+ 2xy + y ^{2}$

=  $2x ^{2}+ 2y ^{2}+ z ^{2}+ 4xy + 2yz + 2xz – 2x ^{2}– 4xy – 2y ^{2}– 2xz – 2yz = z ^{2}$

Vậy $(x + y + z) ^{2}– 2(x + y + z)(x + y) + (x + y) ^{2}= z ^{2}$