How do you simplify $[{(x^{2} - 1)}^{3} (2 x + 1)]$ $[ (x^2-1)^3 (2x+1) ]$ ?

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Answer 1 · 2018-06-25T10:40:00

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

$(x^{2} - 1) (x^{2} - 1) = x^{4} - 2 x^{2} + 1$ $(x^2-1)(x^2-1)=x^4-2x^2+1$

$(x^{4} - 2 x^{2} + 1) (x^{2} - 1) = x^{6} - 2 x^{4} + x^{2} - x^{4} + 2 x^{2} - 1$ $(x^4-2x^2+1)(x^2-1)=x^6-2x^4+x^2-x^4+2x^2-1$

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

$[{(x^{2} - 1)}^{3} (2 x + 1)] = (x^{6} - 3 x^{4} + 3 x^{2} - 1) (2 x + 1)$ $[(x^2-1)^3(2x+1)]=(x^6-3x^4+3x^2-1)(2x+1)$

= $2 x^{7} - 6 x^{5} + 6 x^{3} - 2 x + x^{6} - 3 x^{4} + 3 x^{2} - 1$ $2x^7-6x^5+6x^3-2x+x^6-3x^4+3x^2-1$

= $2 x^{7} + x^{6} - 6 x^{5} - 3 x^{4} + 6 x^{3} + 3 x^{2} - 2 x - 1$ $2x^7+x^6-6x^5-3x^4+6x^3+3x^2-2x-1$

How do you simplify [ (x^2-1)^3 (2x+1) ][(x2−1)3(2x+1)] [ (x^2-1)^3 (2x+1) ]?