Question 1082743

What is the complete binomial expansion for (x^4 − y)^5?
<pre>There will be 6 terms, seeing that the binomial is raised to the 5<sup>rd</sup> power
{{{(a + b)^n = (" "[n]C[0] (a)^(n - 0) (b)^0) + (" "[n]C[1] (a)^(n - 1) (b)^1) + (" "[n]C[2] (a)^(n - 2) (b)^2) + (" "[n]C[3] (a)^(n - 3) (b)^3) + (" "[n]C[4] (a)^(n - 4) (b)^4) + (" "[n]C[5] (a)^(n - 5) (b)^5))}}} ------ Binomial Expansion formula, for 6 terms
{{{(x^4 - y)^5 = (" "[5]C[0] (x^4)^(5 - 0) (- y)^0) + (" "[5]C[1] (x^4)^(5 - 1) (- y)^1) + (" "[5]C[2] (x^4)^(5 - 2) (- y)^2) + (" "[5]C[3] (x^4)^(5 - 3) (-
 y)^3) + (" "[5]C[4] (x^4)^(5 - 4) (- y)^4) + (" "[5]C[5] (x^4)^(5 - 5) (-
 y)^5))}}}
With this being a 5th power binomial, the COMBINATION values in the formula will be the values in the 6th row of PASCAL'S TRIANGLE, which are: 1, 5, 10, 10, 5, and 1.
Therefore, {{{(x^4 - y)^5 = (" "[5]C[0] (x^4)^(5 - 0) (- y)^0) + (" "[5]C[1] (x^4)^(5 - 1) (- y)^1) + (" "[5]C[2] (x^4)^(5 - 2) (- y)^2) + (" "[5]C[3] (x^4)^(5 - 3) (- y)^3) + (" "[5]C[4] (x^4)^(5 - 4) (- y)^4) + (" "[5]C[5] (x^4)^(5 - 5) (- y)^5))}}} becomes: 
           {{{(x^4 - y)^5 = 1(x^4)^(5 - 0) (- y)^0 + 5(x^4)^(5 - 1) (- y)^1 + 10(x^4)^(5 - 2) (- y)^2 + 10(x^4)^(5 - 3) (- y)^3 + 5(x^4)^(5 - 4) (- y)^4 + 1(x^4)^(5 - 5) (- y)^5}}}
           {{{(x^4 - y)^5 = 1(x^4)^5 + 5(x^4)^4 (- y) + 10(x^4)^3y^2 + 10(x^4)^2 (- y)^3 + 5(x^4)^1y^4 + 1(x^4)^0 (- y)^5}}}
           {{{highlight_green((x^4 - y)^5 = x^20 - 5x^16y + 10x^12y^2 - 10x^8y^3 + 5x^4y^4 - y^5)}}}