Question

if x^2+1/x^2=66 ,find x^3-1/x^3

Answer 1

hahakaksjd smakahnaa

Answer 2

$\huge{\underline{\underline{\huge{\bold{AnsweR \: :-}}}}}$

Given :

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

To find : The Value of $x^2 -  \frac{1}{x^3}$

Solution :

$x^2 + \frac{1}{x^2} = 66$

Subtract and add $2 \times x \times\frac{1}{x}$

$x + \frac{1}{x^2} - 2 \times x \times \frac{1}{x} + 2 \times x \times \frac{1}{x} = 66$

$(x - \frac{1}{x})^{2} = 66 - 2 \times x \times \frac{1}{x}$

$(x - \frac{1}{x})^{2} = 64$

Taking root both side :

$x -\frac{1}{x} = 8$

Cubing both side,

$(x - \frac{1}{x})^{3} = {8}^{3}$

${x}^{3} - \frac{1}{ {x}^{3} } - 3 \times x \times \frac{1}{x}(x - \frac{1}{x}) = 512$

${x}^{3} - \frac{1} { {x}^{3} } - 3 \times (8) = 512$

${x}^{3} - \frac{1} { {x}^{3} } = 512 + 24$

${x}^{3} - \frac{1} { {x}^{3} } = 536$

So, the required value is ${x}^{3} - \frac{1} { {x}^{3} } = 536$