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Question

If $$x-\dfrac{1}{x}=5,$$ then find the value of $${x}^{4}+\dfrac{1}{{x}^{4}}$$.


Solution

Given, $$x-\dfrac{1}{x}=5$$

Squaring both sides,
$${\left(x-\dfrac{1}{x}\right)}^{2}={5}^{2}$$

$${\left(x-\dfrac{1}{x}\right)}^{2}=25$$

$$\Rightarrow {x}^{2}+\dfrac{1}{{x}^{2}}-2\times x\times \dfrac{1}{x}=25$$  .......... $$(a-b)^2= a^2-2ab+b^2$$

$$\Rightarrow {x}^{2}+\dfrac{1}{{x}^{2}}-2=25$$

$$\Rightarrow {x}^{2}+\dfrac{1}{{x}^{2}}=25+2=27$$

$$\therefore {x}^{2}+\dfrac{1}{{x}^{2}}=27$$

Squaring both sides,
$${\left({x}^{2}+\dfrac{1}{{x}^{2}}\right)}^{2}={27}^{2}$$

$${x}^{4}+\dfrac{1}{{x}^{4}}+2\times {x}^{2}\times \dfrac{1}{{x}^{2}}=729$$    ............ $$(a+b)^2= a^2+2ab+b^2$$

$${x}^{4}+\dfrac{1}{{x}^{4}}+2=729$$

$${x}^{4}+\dfrac{1}{{x}^{4}}=729-2=727$$

$$\therefore {x}^{4}+\dfrac{1}{{x}^{4}}=727$$

Mathematics

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