# Expand the following: [(a+b^2)(a-b^2)]^5Show the complete solution to explain the answer.

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### 3 Answers

We have to expand [(a+b^2)(a-b^2)]^5

[(a+b^2)(a-b^2)]^5

use (a + b)(a - b) = (a^2 - b^2)

=> [a^2 - b^4]^5

use the binomial theorem (x - y)^5 = -y^5 + 5*x*y^4 - 10*x^2*y^3 + 10*x^3*y^2 - 5*x^4*y + x^5

=> a^10 - 5*a^8*b^4 + 10a^6*b^8 - 10*a^4*b^12 + 5*a^2*b^16 - b^20

**The expansion of [(a+b^2)(a-b^2)]^5 is a^10 - 5*a^8*b^4 + 10a^6*b^8 - 10*a^4*b^12 + 5*a^2*b^16 - b^20**

Well, this polynomial first has a difference of squares inside

Mutiplying them out

(a+b^2)(a-b^2)=a^2-b^4

The whole polynomial becomes

(a^2-b^4)^5

Using Pascal's Triangle and the Binomial Theorem:

1 first degree

1 2 1 second degree

1 3 3 1 third degree

1 4 6 4 1 fourth degree

1 5 10 10 5 1 fifth degree

so the polynomial (a^2-b^4))^5= (a^2))^5- 5((a^2)^4))b^4 + 10 ((a^2)^3)((b^4)^2)-10((a^2)^2)((b^4)^3)+5a^2((b^4)^4)+(b^4)^5

simplify this huge expression

=** a^10-5(a^8)*(b^4)+10(a^6)*(b^8)-10(a^4)*(b^12)+5(a^2)*(b^16)-b^20**

*Actually, for any polynomial to the exponent 5*

*(x+y)^5=x^5+5(x^4)*y+10(x^3)*(y^2)+10(x^2)*(y^3)+*

*5x*(y^4)+y^5*

*You could see, the x-exponent is decreasing by one every term, and the y exponent is increasing by one in every term.*

*The coefficents are given by the Pascal Triangle.*

The product within brackets returns the difference of two squares:

(a+b^2)(a-b^2) = a^2 - b^4

Now, we'll develop the binomial using binomial theorem:

(a^2 - b^4)^5 =C(5,0)*(a^2)^5 - C(5,1)*(a^2)^4*b^4 + C(5,2)(a^2)^3*(b^4)^2 - C(5,3)*(a^2)^2*(b^4)^3 + C(5,4)*(a^2)*(b^4)^4 - C(5,5)*(b^4)^5

We'll use the factorial identity to calculate C(n,k)

C(n,k) = n!/k!(n-k)!

C(5,0) = 1

C(5,1) = 5

C(5,2) = 5(5-1)/2 = 10

C(5,3) = 10

C(5,4) = 5

C(5,5) = 1

**Therefore, the expansion of the given binomial is: (a^2 - b^4)^5 = a^10 - 5(a^8)*(b^4) + 10(a^6)*(b^8) - 10(a^4)*(b^12) + 5(a^2)*(b^16) - (b^20)**