Question

Factorize x square + 1/ x square+2-3x-3/x​

Answer 1

Answer:

$x^{2}+\frac{1}{x^{2}}+2-3x-\frac{3}{x}=\left(x+\frac{1}{x}\right)[\left(x+\frac{1}{x}-3\right)$

Step-by-step explanation:

$x^{2}+\frac{1}{x^{2}}+2-3x-\frac{3}{x}\\=(x^{2}+\frac{1}{x^{2}}+2\times x\times \frac{1}{x})-3x-\frac{3}{x}\\=\left(x+\frac{1}{x}\right)^{2}-3\left(x+\frac{1}{x}\right)\\=\left(x+\frac{1}{x}\right)[\left(x+\frac{1}{x}-3\right)$

Therefore,

$x^{2}+\frac{1}{x^{2}}+2-3x-\frac{3}{x}=\left(x+\frac{1}{x}\right)[\left(x+\frac{1}{x}-3\right)$

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Answer 2

Answer:

Step-by-step explanation: