Find the square root of the polynomial using factorisation method: $y^{8} + \frac{1}{y^{8}} - 2$ .

∣ ∣ ∣ y^{3} + \frac{1}{y^{3}} ∣ ∣ ∣

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∣ ∣ ∣ y^{2} + \frac{1}{y^{2}} ∣ ∣ ∣

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∣ ∣ ∣ y^{4} + \frac{1}{y^{4}} ∣ ∣ ∣

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y^{4} + \frac{1}{y^{4}}

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Solution

The correct option is B $∣ ∣ ∣ y^{4} + \frac{1}{y^{4}} ∣ ∣ ∣$
$\sqrt{y^{8} + \frac{1}{y^{8}} - 2}$
$=$ $\sqrt{y^{8} + \frac{1}{y^{8}} - 2}$
$=$ $\sqrt{(y^{4} + \frac{1}{y^{4}})^{2}}$ $[(y^{4} + \frac{1}{y^{4}})^{2} = (y^{4})^{2} + 2 (y^{4}) (\frac{1}{y^{4}}) + (\frac{1}{y^{4}})^{2}]$
$=$ $∣ ∣ ∣ y^{4} + \frac{1}{y^{4}} ∣ ∣ ∣$

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