Question
Find the following:
$\left(x^2-1\right)\left(x^4+x^2+1\right)$
$\left(x^2-1\right)\left(x^4+x^2+1\right)$
✓
Answer
We have,
$\left(x^2-1\right)\left(x^4+x^2+1\right)$
$=\left(x^2-1\right)\left[\left(x^2\right) 2+1 \times x^2+1^2\right]$
$=\left(x^2\right)^3-(1)^3\left[\because a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=x^6-1$
$\therefore\left(x^2-1\right)\left(x^4+x^2+1\right)=x^6-1$
$\left(x^2-1\right)\left(x^4+x^2+1\right)$
$=\left(x^2-1\right)\left[\left(x^2\right) 2+1 \times x^2+1^2\right]$
$=\left(x^2\right)^3-(1)^3\left[\because a^3-b^3=(a-b)\left(a^2+a b+b^2\right)\right]$
$=x^6-1$
$\therefore\left(x^2-1\right)\left(x^4+x^2+1\right)=x^6-1$
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