Question

x²+1/x²=7,find x³+1/x³

Answer 1

hey dear

here is your answer

Solution

x^2 + 1/x^2 = 7

to find x^3 + 1/x^3 = ?

( x + 1 / x) ^2 (- 2 *x * 1/x) =7

(x +1 /x) ^2 -2 =7

(x + 1/x )^2 = 7+2

( x+1/x)^2 = 9

take square root on both the side

( x+1/x ) =3

( x^3 + 1/x3 ) = ( x + 1/x )^3. -3 *x * 1/x. ( x +1/x)

(x + 1/x )^3 - 3 ( x+1/x)

(3)^3 - 3 (3)

27 -9

=18

hence x^3 +1/x3 = 18

hence your answer is 18

hope it helps

thank you

Answer 2

Hey !!

Here is your answer.. ⬇⬇⬇

${x}^{2} + { (\frac{1}{x}) }^{2} = ({x + \frac{1}{x} )}^{2} \\ \\ {(x + \frac{1}{x}) }^{2} = {x}^{2} + 2 + {( \frac{1}{x} )}^{2} \\ \\ {(x + \frac{1}{x}) }^{2} = 7 + 2 \\ \\ {(x + \frac{1}{x} )}^{2} = 9 \\ \\ x + \frac{1}{x} = \sqrt{9} \\ \\ x + \frac{1}{x} = + 3 and -3 \\ \\ now... \\ \\ {(x + \frac{1}{x} )}^{3} = {x}^{3} + { (\frac{1}{x} )}^{3} + 3(x + \frac{1}{x} ) \\ \\ ({3})^{3} = {x}^{3} + { (\frac{1}{x} )}^{3} + 3(3) \\ 27 = {x}^{3} + { (\frac{1}{x} )}^{3} + 9 \\ \\ {x}^{3} + { (\frac{1}{x}) }^{3} = 27 - 9 \\ \\ {x}^{3} + { (\frac{1}{x} )}^{3} = 18$

in 2nd case
x + 1/x = -3

( x + 1/x )^3 = x^3 + 1/x^3 + 3( x + 1/x )

-27 = x^3 + 1/x^3 + 3 ( -3 )

-27 = x^3 + 1/x^3 -9

x^3 + 1/x^3 = -9 + 27

x^3 + 1/x^3 = 18

HOPE IT HELPS...

THANKS ^-^