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Proving Polynomial Identities

Proving Polynomial Identities 1.6 - Solution

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a
The difference of squares has the form (ab)(a+b)=a2b2 (a - b)(a + b) = a^2 - b^2 and we see that our expression has the same form, where a=pa=p and b=64b = 64.
(p64)(p+64)\left(p-64\right)\left(p+64\right)
p2642p^2 - 64^2
p24096p^2 - 4096
b
The difference of squares has the form (a+b)(ab)=a2b2 (a + b)(a - b) = a^2 - b^2 and we see that our expression has the same form, with a=xa=x and b=2yb = 2y.
(x+2y)(x2y)(x+2y)(x-2y)
x2(2y)2x^2 - (2y)^2
x222y2x^2 - 2^2y^2
x24y2x^2 - 4y^2
c
We proceed in the same way to rewrite the expression with the difference of two squares. Note that the square of x2x^2 is (x2)2.(x^2)^2.
(x28)(x2+8)\left(x^2-8\right)\left(x^2+8\right)
(x2)282\left(x^2\right)^2 - 8^2
x482x^4 - 8^2
x464x^4 - 64