Question

if x² + 1/x² = 38 , then find x³ - 1/x³​

Answer 1

Given,

equation given is

$x^{2} +\frac{1}{x^{2} }$  =38                                                    (1)

To Find,

the value of $x^{3} -\frac{1}{x^{3} }$

Solution,

we will use the identity $(x-\frac{1}{x} )^{2}$ = $x^{2} +\frac{1}{x^{2} } -2$ to find the required value.

so,

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

 $(x-\frac{1}{x}) ^{2} = 38- 2$                                       (using (1))

$(x-\frac{1}{x}) ^{2} =36$

$x-\frac{1}{x} = \sqrt{36}$

$x-\frac{1}{x} = 6$                                                           (2)

now, we will apply identity $(x-\frac{1}{x} )^{3} = x^{3} -\frac{1}{x^{3} } -3(x-\frac{1}{x} )$

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

$(6)^{3} = x^{3} - \frac{1}{x^{3} } -3(6)$                                    ( using (2))

216 = $x^{3} -\frac{1}{x^{3} } -18$

$x^{3}-\frac{1}{x^{3} } = 216+18 = 234$

Hence, the value  $x^{3}-\frac{1}{x^{3} }$ is 234.