What is expand (x+1)(x-1)(x^2-1)^2 ?

The solution to expand (x+1)(x-1)(x^2-1)^2 is x^6-3x^4+3x^2-1

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expand (x+1)(x-1)(x^2-1)^2

expand

(x + 1) (x - 1) (x^{2} - 1)^{2}

x^{6} - 3 x^{4} + 3 x^{2} - 1

Solution steps

(x + 1) (x - 1) (x^{2} - 1)^{2}

= (x^{2} - 1)^{2} (x + 1) (x - 1)

Apply Difference of Two Squares

(x + 1) (x - 1) : x^{2} - 1

= (x^{2} - 1)^{2} (x^{2} - 1)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

(x^{2} - 1)^{2} (x^{2} - 1) = (x^{2} - 1)^{2 + 1}

= (x^{2} - 1)^{2 + 1}

Add the numbers:

2 + 1 = 3

= (x^{2} - 1)^{3}

Apply Perfect Cube Formula:

(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}

(x^{2} - 1)^{3} = (x^{2})^{3} - 3 (x^{2})^{2} \cdot 1 + 3 x^{2} \cdot 1^{2} - 1^{3}

= (x^{2})^{3} - 3 (x^{2})^{2} \cdot 1 + 3 x^{2} \cdot 1^{2} - 1^{3}

Simplify

(x^{2})^{3} - 3 (x^{2})^{2} \cdot 1 + 3 x^{2} \cdot 1^{2} - 1^{3} : x^{6} - 3 x^{4} + 3 x^{2} - 1

= x^{6} - 3 x^{4} + 3 x^{2} - 1

(- 2) (- 5)

\frac{7 - 5 i}{5 + 9 i}

15 (6 - g)

(x^{2} + 3 x + 3) - (7 x^{2} - x - 1)

- 3 x + 9 x

What is expand (x+1)(x-1)(x^2-1)^2 ?
The solution to expand (x+1)(x-1)(x^2-1)^2 is x^6-3x^4+3x^2-1