What is expand (p^2-1)(p^2+1)(p^4+1)(p^8+1) ?

The solution to expand (p^2-1)(p^2+1)(p^4+1)(p^8+1) is p^{16}-1

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expand (p^2-1)(p^2+1)(p^4+1)(p^8+1)

expand

(p^{2} - 1) (p^{2} + 1) (p^{4} + 1) (p^{8} + 1)

p^{16} - 1

Solution steps

(p^{2} - 1) (p^{2} + 1) (p^{4} + 1) (p^{8} + 1)

Expand

(p^{2} - 1) (p^{2} + 1) (p^{4} + 1) (p^{8} + 1) : (p^{4} - 1) (p^{4} + 1) (p^{8} + 1)

= (p^{4} - 1) (p^{4} + 1) (p^{8} + 1)

Apply Difference of Two Squares

(p^{4} - 1) (p^{4} + 1) : p^{8} - 1

= (p^{8} - 1) (p^{8} + 1)

Apply Difference of Two Squares

(p^{8} - 1) (p^{8} + 1) : p^{16} - 1

= p^{16} - 1

l {\frac{( 1 - cos ( t ) )}{t}}

\frac{3}{4} x + \frac{1}{4} (x - 6) + 28.4

8 a^{3} - 36 a^{2} + 54 a - 27

\frac{12 x ^{5} \cdot y ^{2}}{3 \cdot x ^{2} \cdot y}

26 x (- 48) + (- 48) \cdot x (- 36)

What is expand (p^2-1)(p^2+1)(p^4+1)(p^8+1) ?
The solution to expand (p^2-1)(p^2+1)(p^4+1)(p^8+1) is p^{16}-1