Преобразуйте выражение:1) (b^n-b) ^22) (x^m-x) ^23) (c^k+1+c^k) ^24) (a^m+b^n) ^25)...

Решите задачу:

$1) (b^n-b)^2=(b-(-b))^2=(b^n+b)^2=b^{2n}+2b^{n+1}+b^2 2) (x^m-x)^2=(x^m)^2-2x^mx+x^2=x^{2m}-2x^{m+1}+x^2 3) (c^k+1+c^k)^2=(2c^k+1)^2=4c^{2k}+4c^k+1 4) (a^m+b^n)^2=(a^m)^2+2a^mb^n+(b^n)^2 5) (3/4a^2-0,5b^3)^2=(3/4a^2-1/2b^3)^2=9/16a^4-3/4a^2b^3+1/4b^6 6) (5/3x^2+0,6y^4)^2=(5/3x^2+3/5y^4)^2=25/9x^4+2x^2y^4+9/25y^8$