Question
Simplify:
$ (2 x-1)(2 x+1)\left(4 x^2+1\right)\left(16 x^4+1\right) $
$ (2 x-1)(2 x+1)\left(4 x^2+1\right)\left(16 x^4+1\right) $
✓
Answer
To simplify, we will proceed as follows:
$ (2 x-1)(2 x+1)\left(4 x^2+1\right)\left(16 x^4+1\right) $
$ =\left((2 x)^2-1^2\right)\left(4 x^2+1\right)\left(16 x^4+1\right)\left[\because(a+b)(a-b)=a^2-b^2\right] $
$ =\left(4 x^2-1\right)\left(4 x^2+1\right)\left(16 x^4+1\right) $
$ =\left\{\left(4 x^2\right)^2-\left(1^2\right)^2\right\}\left(16 x^4+1\right)\left[\because(a+b)(a-b)=a^2-b^2\right] $
$ =\left(16 x^4-1\right)\left(16 x^4+1\right)$
$ =\left(16 x^4\right)^2-1^2\left[\because(a+b)(a-b)=a^2-b^2\right] $
$ =256 x^8-1 $
$ (2 x-1)(2 x+1)\left(4 x^2+1\right)\left(16 x^4+1\right) $
$ =\left((2 x)^2-1^2\right)\left(4 x^2+1\right)\left(16 x^4+1\right)\left[\because(a+b)(a-b)=a^2-b^2\right] $
$ =\left(4 x^2-1\right)\left(4 x^2+1\right)\left(16 x^4+1\right) $
$ =\left\{\left(4 x^2\right)^2-\left(1^2\right)^2\right\}\left(16 x^4+1\right)\left[\because(a+b)(a-b)=a^2-b^2\right] $
$ =\left(16 x^4-1\right)\left(16 x^4+1\right)$
$ =\left(16 x^4\right)^2-1^2\left[\because(a+b)(a-b)=a^2-b^2\right] $
$ =256 x^8-1 $
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