Question

Plz solve this question..!!
If x+1/x=9 and x-1/x=6,then find the value of x4-1/x4.

Answer 1

Given: $x+\dfrac{1}{x} =9 \text { and } x-\dfrac{1}{x} =6$

To find: $x^4-\dfrac{1}{x^4}$

Step-by-step explanation:

As we know the algebraic identities

$(a+b)^2=a^2+b^2+2ab---- I$

1) As we have

$x+\dfrac{1}{x} =9 \\\\\text {Squaring both sides and applying algeraric identity I we get }\\\\\Rightarrow(x+\dfrac{1}{x})^2 =9 ^2\\\\\Rightarrow x^2+\dfrac{1}{x^2} +2x\times \dfrac{1}{x} =81\\\\\Rightarrow x^2+\dfrac{1}{x^2} +2=81\\\\\Rightarrow x^2+\dfrac{1}{x^2} =79$

Now

$(a+b)(a-b)=a^2-b^2$

Therefore

$x^4-\dfrac{1}{x^4}\\\\= (x^2+\dfrac{1}{x^2} )(x^2-\dfrac{1}{x^2})\\\\=(x^2+\dfrac{1}{x^2})(x+\dfrac{1}{x})(x-\dfrac{1}{x})$

Substituting the values

$= 79 \times 9\times 6 = 4266$

Hence the value of $x^4-\dfrac{1}{x^4}$   is 4266