Question
Simplify the following products:
$(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)$
$(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)$
✓
Answer
In the given problem, we have to find product of
$(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)$
taking x as common factor $\text{x}(\text{x}^2 -3\text{x}-1)(\text{x}^2-3\text{x} + 1)$
$(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)\\=\Big[\text{x}\big(\text{x}^2-3\text{x}-1\big)\big(\text{x}^2-3\text{x}+1\big)\Big]$
$=\text{x}\Big[\big\{\big(\text{x}^2-3\text{x}\big)-1\big\}\big\{\big(\text{x}^2-3\text{x}\big)+1\big\}\Big]$
We shall use the identity $(\text{x}-\text{y})(\text{x}+\text{y})=\text{x}^2-\text{y}^2$
$\big(\text{x}^3-3\text{x}^2-\text{x}\big)\big(\text{x}^2-3\text{x}+1\big)\\=\text{x}\Big[\big(\text{x}^2-3\text{x}\big)^2-1^2\Big]$
$=\text{x}\big(\text{x}^4-6\text{x}^3+9\text{x}^2-1\big)$
$=\text{x}^5-6\text{x}^4+9\text{x}^3-\text{x}$
Hence the value of $(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)$ is $\text{x}^5-6\text{x}^4+9\text{x}^3-\text{x}.$
$(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)$
taking x as common factor $\text{x}(\text{x}^2 -3\text{x}-1)(\text{x}^2-3\text{x} + 1)$
$(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)\\=\Big[\text{x}\big(\text{x}^2-3\text{x}-1\big)\big(\text{x}^2-3\text{x}+1\big)\Big]$
$=\text{x}\Big[\big\{\big(\text{x}^2-3\text{x}\big)-1\big\}\big\{\big(\text{x}^2-3\text{x}\big)+1\big\}\Big]$
We shall use the identity $(\text{x}-\text{y})(\text{x}+\text{y})=\text{x}^2-\text{y}^2$
$\big(\text{x}^3-3\text{x}^2-\text{x}\big)\big(\text{x}^2-3\text{x}+1\big)\\=\text{x}\Big[\big(\text{x}^2-3\text{x}\big)^2-1^2\Big]$
$=\text{x}\big(\text{x}^4-6\text{x}^3+9\text{x}^2-1\big)$
$=\text{x}^5-6\text{x}^4+9\text{x}^3-\text{x}$
Hence the value of $(\text{x}^3 -3\text{x}^2-\text{x})(\text{x}^2-3\text{x} + 1)$ is $\text{x}^5-6\text{x}^4+9\text{x}^3-\text{x}.$
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