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