Question

If x-1/x=9, find the value of x²+1/x²​

Answer 1

Hence the answer is 79

hope it will help you

Answer 2

Given :

$x - \dfrac{1}{x} = 9$

To find :

${x}^{2} + \dfrac{1}{{x}^{2} }$

Solution :

$\implies {(x - \dfrac{1}{x} )}^{2} = {x}^{2} + { \dfrac{1}{x} }^{2} -2. x. \dfrac{1}{x}$

$\implies {(x - \dfrac{1}{x} )}^{2} = {x}^{2} + { \dfrac{1}{x} }^{2} -2$

We know that :-

$x - \dfrac{1}{x} = 9$

$\implies {9}^{2} = {x}^{2} + { \dfrac{1}{x} }^{2} -2$

we know that :- 9²=81

$\implies 81 = {x}^{2} + { \dfrac{1}{x} }^{2} -2$

$\implies 81 + 2= {x}^{2} + { \dfrac{1}{x} }^{2}$

$\implies 83= {x}^{2} + { \dfrac{1}{x} }^{2}$

Answer : 83

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$\implies{(a+b)^2 = a^2 + b^2 + 2ab}$

$\implies{(a-b)^2 = a^2 + b^2 - 2ab}$

$\implies{(a+b)^3 = a^3 + b^3 + 3ab(a + b)}$

$\implies{(a-b)^3 = a^3 - b^3 - 3ab(a-b)}$

$\implies{(a^3+b^3)= (a+b)(a^2 - ab + b^2)}$

$\implies{(a^3-b^3)= (a-b)(a^2 + ab + b^2)}$

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